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Sequences and Series: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 118 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. A sum, relabeled and shifted . 12 points. Question 1 of 10.

Consider the sum $4^2+5^2+6^2+7^2+8^2+9^2+10^2$ .
1. Part A.
  Write this sum in sigma notation, using $j$ as the index and starting the index at $j=4$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Evaluate $\displaystyle\sum_{j=4}^{10} j^2$ , and state how many terms it has.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Rewrite the same sum with a new index $p$ , substituting $j=p+4$ throughout (including in the limits, so $p=0$ corresponds to the first term). Confirm your rewritten sum evaluates to the same total, and explain why this kind of substitution is guaranteed to preserve the value of a sum.
  
  Carry your own answer forward Use your total from part B, whatever it came out to be, to check that the rewritten sum matches it.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$\displaystyle\sum_{j=4}^{10} j^2$

Part B

The sum has $7$ terms, and $\displaystyle\sum_{j=4}^{10} j^2 = 371$ .

Part C

$\sum_{p=0}^{6}(p+4)^2=371$ , matching part B; shifting the index just relabels the same seven terms in the same order, so the total is unchanged.

Worked solution

Part A

Each term is an integer squared, and that integer counts upward by one from term to term, starting at $4$ and ending at $10$ . So the term at index $j$ is $j^2$ , and the limits match where the base starts and stops.

$4^2+5^2+6^2+7^2+8^2+9^2+10^2 = \sum_{j=4}^{10} j^2.$

Part B

Both limits are included, so the term count is $10-4+1=7$ .

$16+25+36+49+64+81+100 = 371.$

Part C

Substituting $j=p+4$ turns $j=4$ into $p=0$ and $j=10$ into $p=6$ , and the summand becomes $(p+4)^2$ .

$\sum_{p=0}^{6}(p+4)^2 = 4^2+5^2+\cdots+10^2 = 371,$

since at $p=0,1,\ldots,6$ the quantity $p+4$ runs through exactly $4,5,\ldots,10$ , the same seven values as before in the same order. A sum never depends on what its index is called or where its counting starts, only on which list of values gets added; relabeling that list changes nothing about the list itself.

In one line

$4^2+5^2+\cdots+10^2=\displaystyle\sum_{j=4}^{10}j^2$ , which has $7$ terms and evaluates to $371$ . Substituting $j=p+4$ gives $\displaystyle\sum_{p=0}^{6}(p+4)^2$ , which also totals $371$ , because the substitution only relabels the same seven values in the same order.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Identifies $j^2$ as the summand, matching the pattern of the given terms. . Worth 2 points.

Sets the lower limit at $4$ and the upper limit at $10$ , matching the first and last given terms. . Worth 2 points.

Part B 4 points

Reports the term count, correctly computed from both limits of the sum. . Worth 1 point.

Evaluates the summand at every index from $4$ to $10$ and adds all seven results correctly. . Worth 2 points.

Reports the term count and the sum as two separate, clearly labeled quantities. . Worth 1 point.

Part C 4 points

Correctly substitutes $j=p+4$ into the summand and both limits, and confirms the rewritten sum matches the total from part B. . Worth 2 points.

Explains that the substitution merely relabels the same list of values in the same order, so the sum's value cannot change. . Worth 2 points. needs an explanation, not just an answer
2. A shrinking sequence and how soon it drops below a bound . 12 points. Question 2 of 10.

A geometric sequence begins $18, 6, 2, \tfrac23, \ldots$
1. Part A.
  Confirm the common ratio, write the explicit formula for $a_n$ , and use it to find $a_8$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Find $S_8$ , the sum of the first eight terms, as a fraction in lowest terms.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Find the smallest number of terms $n$ after which a term of this sequence first drops below $0.01$ . Then explain why, once that happens, no later term can climb back above $0.01$ .
  
  Carry your own answer forward Use the explicit formula you wrote in part A to set up the inequality.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$r=\tfrac13$ ; $a_n=18\left(\tfrac13\right)^{n-1}$ ; $a_8=\tfrac{2}{243}$ .

Part B

$S_8=\tfrac{6560}{243}$ .

Part C

The smallest such $n$ is $8$ ; since $0<r<1$ here, each term is smaller than the last, so once a term drops below $0.01$ , every later term, being smaller still, stays below it too.

Worked solution

Part A

Dividing consecutive terms gives $r=\tfrac{6}{18}=\tfrac13$ , confirmed by $\tfrac{2}{6}=\tfrac13$ . With $a_1=18=2\cdot3^2$ ,

$a_n = 18\left(\frac13\right)^{n-1}, \qquad a_8 = \frac{18}{3^{\,7}} = \frac{2\cdot3^2}{3^{\,7}}.$

Two of the seven factors of $3$ in the denominator cancel against the $3^2$ on top, leaving

$a_8 = \frac{2}{3^5} = \frac{2}{243}.$

Part B

Apply the finite sum formula with $a=18$ , $r=\tfrac13$ , $n=8$ .

$S_8 = \frac{18\left(1-\left(\tfrac13\right)^8\right)}{1-\tfrac13} = \frac{18\left(1-\tfrac{1}{6561}\right)}{\tfrac23} = 27\left(\frac{6560}{6561}\right) = \frac{6560}{243}.$

Part C

Solve $18\left(\tfrac13\right)^{n-1} < 0.01$ for $n$ .

$\left(\frac13\right)^{n-1} < \frac{0.01}{18} \quad\Longrightarrow\quad (n-1)\log\!\left(\frac13\right) < \log\!\left(\frac{0.01}{18}\right).$

Since $\log\!\left(\tfrac13\right)<0$ , dividing reverses the inequality: $n-1 > 6.82\ldots$ , so $n-1=7$ and $n=8$ . Checking, $a_7=\tfrac{2}{81}\approx0.0247>0.01$ while $a_8=\tfrac{2}{243}\approx0.0082<0.01$ .

With $0<r=\tfrac13<1$ , every term is exactly $\tfrac13$ of the one before it, so the sequence strictly decreases forever. A term that has already fallen under $0.01$ can therefore never be exceeded by a later, smaller term.

In one line

For $18,6,2,\tfrac23,\ldots$ : $r=\tfrac13$ , $a_n=18\left(\tfrac13\right)^{n-1}$ , $a_8=\tfrac{2}{243}$ , and $S_8=\tfrac{6560}{243}$ . A term first drops below $0.01$ at $n=8$ , and because $0<r<1$ makes every later term smaller still, no term can climb back above $0.01$ afterward.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Finds the ratio by dividing consecutive terms and checks it against a second pair. . Worth 1 point.

Writes the explicit formula and uses the exponent $7$ , not $8$ , when substituting $n=8$ . . Worth 2 points.

Reduces $a_8$ to lowest terms. . Worth 1 point.

Part B 4 points

Selects the finite-sum formula with the correct $a$ , $r$ , and $n$ . . Worth 1 point.

Carries the fraction arithmetic through correctly to $S_8=\tfrac{6560}{243}$ . . Worth 2 points.

Reports the result reduced to lowest terms. . Worth 1 point.

Part C 4 points

Sets up and solves the correct inequality with logarithms, handling the reversal from dividing by a negative logarithm. . Worth 2 points.

Explains, from $0<r<1$ , why the sequence strictly decreases and so cannot climb back above the bound once it has dropped below it. . Worth 2 points. needs an explanation, not just an answer
3. A rule that reaches back one step, and the shortcut hiding inside it . 11 points. Question 3 of 10.

A sequence is defined recursively by $k_1 = 11$ and $k_n = k_{n-1} + 6$ for $n \ge 2$ .
1. Part A.
  Find $k_2$ , $k_3$ , and $k_4$ , and state the sequence's common difference.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Write the explicit formula for $k_n$ in simplified form, and use it to find $k_{15}$ .
  
  Carry your own answer forward Use the seed and common difference you found in part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  A classmate argues that because $k_n$ was defined recursively, it cannot also have an explicit formula like the one in part B, since 'recursive' and 'explicit' name two completely different kinds of rule. Explain what is wrong with this reasoning, using this sequence, its seed, and part B's formula as your example.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$k_2=17$ , $k_3=23$ , $k_4=29$ ; the common difference is $d=6$ .

Part B

$k_n=6n+5$ ; $k_{15}=95$ .

Part C

Both descriptions can be true of the same sequence: 'recursive' and 'explicit' describe two different ways of WRITING a rule, not two different sequences, and $k_n=6n+5$ can be checked to satisfy the recursive rule at every step.

Worked solution

Part A

Build each term from the one before it, using the single seed $k_1=11$ .

$k_2 = 11+6=17, \qquad k_3=17+6=23, \qquad k_4=23+6=29.$

Every consecutive difference is $6$ , so $d=6$ .

Part B

Substitute $k_1=11$ and $d=6$ and simplify.

$k_n = 11+(n-1)6 = 6n+5, \qquad k_{15} = 6(15)+5 = 95.$

Part C

'Recursive' and 'explicit' are two different formats for describing a sequence, not two different sequences competing for the same terms. This sequence needed only one seed, $k_1=11$ , because its recursive rule looks back a single step; that one seed, together with the step rule, pins down a unique list of terms. The explicit formula $k_n=6n+5$ from part B is not a rival rule, it is a closed-form description of that exact same list: it reproduces the seed, and it satisfies the recursive step at every $n$ .

$k_1 = 6(1)+5 = 11, \qquad (6n+5)-\bigl(6(n-1)+5\bigr) = 6 \text{ for every } n.$

A sequence is the list of terms itself, and both a recursive rule and an explicit formula are just two languages for naming the same list.

In one line

With seed $k_1=11$ and $k_n=k_{n-1}+6$ : $k_2=17$ , $k_3=23$ , $k_4=29$ , $d=6$ , and $k_n=6n+5$ gives $k_{15}=95$ . A recursive rule and an explicit formula are two different descriptions of the same sequence, not two different sequences, and $k_n=6n+5$ can be checked to satisfy the recursive rule at every step.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Builds $k_2$ , $k_3$ , $k_4$ in order from the single seed $k_1$ . . Worth 2 points.

Confirms the common difference is constant across the computed terms and states $d=6$ . . Worth 1 point.

Part B 4 points

Substitutes correctly and simplifies to $k_n=6n+5$ . . Worth 2 points.

Evaluates the simplified formula at $n=15$ to get $k_{15}=95$ . . Worth 1 point.

Reports the result as the value of the 15th term, not the position number itself. . Worth 1 point.

Part C 4 points

States that a recursive rule and an explicit formula are two different ways of describing the same sequence, not two competing sequences. . Worth 2 points. needs an explanation, not just an answer

Supports the claim by checking that $k_n=6n+5$ reproduces the seed and satisfies the recursive step, or gives an equivalent concrete check. . Worth 2 points.
4. Which two partial sums actually isolate a range . 13 points. Question 4 of 10.

An arithmetic sequence has $a_1 = -8$ and $d = 5$ .
1. Part A.
  Find $a_{19}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Find the sum of the terms from $a_9$ through $a_{19}$ , inclusive.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A classmate tries to find the sum from $a_9$ through $a_{19}$ a different way, by computing $S_{19}$ and $S_9$ and subtracting them, and gets an answer that does not match part B. Identify the classmate's error, and state which two partial sums should have been subtracted instead.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
The answer

Part A

$a_{19}=82$ .

Part B

The sum from $a_9$ through $a_{19}$ is $627$ .

Part C

Subtracting $S_{19}-S_9$ removes $a_9$ itself, which belongs in the range; the correct pair is $S_{19}-S_8$ , since $S_8$ stops just before $a_9$ and leaves the whole range $a_9$ through $a_{19}$ intact.

Worked solution

Part A

Reaching term $19$ from term $1$ takes $18$ steps.

$a_{19} = -8+18(5) = -8+90 = 82.$

Part B

This range has $19-9+1=11$ terms, running from $a_9=-8+8(5)=32$ to $a_{19}=82$ .

$S = \frac{11}{2}(32+82) = \frac{11}{2}(114) = 627.$

Part C

$S_9$ already includes the term $a_9$ , so subtracting $S_{19}-S_9$ removes $a_9$ along with $a_1$ through $a_8$ , throwing away a term the range was supposed to keep. The correct partial sum to subtract is $S_8$ , the running total through the term just before the range begins:

$S_{19}-S_8 \quad\text{leaves exactly } a_9, a_{10}, \ldots, a_{19} \text{ behind.}$

Computing $S_8=\tfrac{8}{2}(-8+27)=4(19)=76$ and $S_{19}=\tfrac{19}{2}(-8+82)=\tfrac{19}{2}(74)=703$ confirms it: $703-76=627$ , matching part B.

In one line

With $a_1=-8$ and $d=5$ : $a_{19}=82$ , and the sum from $a_9$ through $a_{19}$ is $627$ . Subtracting $S_{19}-S_9$ wrongly removes $a_9$ itself; the correct pair is $S_{19}-S_8=703-76=627$ , since $S_8$ stops just before the range begins.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Uses $19-1=18$ steps in the explicit formula, not $19$ . . Worth 1 point.

Computes $-8+18(5)=82$ correctly. . Worth 2 points.

Reports the result as the value of the 19th term, not the position number itself. . Worth 1 point.

Part B 5 points

Counts the range from $a_9$ through $a_{19}$ inclusively as $11$ terms and finds $a_9=32$ . . Worth 2 points.

Applies the sum formula to the $11$ -term range correctly to get $627$ . . Worth 2 points.

Reports the result as the total for the stated range, not as $S_{19}$ or any other full running total. . Worth 1 point.

Part C 4 points

Identifies that $S_9$ already includes $a_9$ , so subtracting it removes a term that belongs in the range. . Worth 2 points.

States that $S_{19}-S_8$ is the correct pair, and confirms it reproduces the total from part B. . Worth 2 points.
5. A finite total, and how close it already sits to the infinite one . 12 points. Question 5 of 10.

Two geometric series share the same first term and ratio: $a_1=48$ and $r=\tfrac14$ .
1. Part A.
  Find $S_6$ , the sum of the first six terms, as a fraction in lowest terms.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Now suppose the series continues forever. Find its sum $S$ , and compute exactly how far $S_6$ from part A falls short of $S$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Explain why $S_6$ , or any partial sum of this series, could never overshoot $S$ , for any series with $a_1>0$ and $0<r<1$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 3 points
The answer

Part A

$S_6=\tfrac{4095}{64}$ .

Part B

$S=64$ ; $S-S_6=\tfrac{1}{64}$ .

Part C

It can never overshoot: $S-S_n=S\,r^{\,n}$ , and with $S>0$ and $0<r<1$ this quantity is always positive, so $S_n<S$ at every single $n$ , however many terms are added.

Worked solution

Part A

Apply the finite sum formula with $a=48$ , $r=\tfrac14$ , $n=6$ .

$S_6 = \frac{48\left(1-\left(\tfrac14\right)^6\right)}{1-\tfrac14} = 64\left(1-\frac{1}{4096}\right) = \frac{4095}{64}.$

Part B

Since $\left|\tfrac14\right|<1$ , the series converges.

$S = \frac{48}{1-\tfrac14} = \frac{48}{\tfrac34} = 64.$

The exact gap is $S-S_6=S\,r^6=64\left(\tfrac14\right)^6=\dfrac{64}{4096}=\dfrac{1}{64}$ , which also equals $64-\tfrac{4095}{64}=\tfrac{1}{64}$ directly.

Part C

The exact gap between any partial sum and the true sum is $S-S_n=S\,r^{\,n}$ . With $a_1>0$ and $0<r<1$ , the sum $S=\dfrac{a_1}{1-r}$ is itself positive, and $r^n$ is positive for every positive integer $n$ since a positive number raised to any power stays positive.

$S-S_n = S\,r^{\,n} > 0 \quad\Longrightarrow\quad S_n < S \text{ for every } n.$

So every partial sum falls strictly short of $S$ ; none of them can ever reach or pass it, however many terms are added.

In one line

For $a_1=48$ , $r=\tfrac14$ : $S_6=\tfrac{4095}{64}$ , and the infinite sum is $S=64$ , with $S-S_6=\tfrac{1}{64}$ exactly. Since $S-S_n=S\,r^{\,n}$ is always positive when $a_1>0$ and $0<r<1$ , every partial sum of such a series falls strictly short of $S$ and can never overshoot it.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Selects the finite-sum formula with the correct $a$ , $r$ , $n$ . . Worth 1 point.

Carries the fraction arithmetic through correctly to $S_6=\tfrac{4095}{64}$ . . Worth 2 points.

Reports the result reduced to lowest terms. . Worth 1 point.

Part B 5 points

Confirms $|r|<1$ and applies the infinite sum formula to get $S=64$ . . Worth 2 points.

Computes the exact gap $S-S_6=\tfrac{1}{64}$ , by the error formula or by direct subtraction. . Worth 2 points.

Reports $S$ as the number the partial sums close in on, distinct from the gap $S-S_6$ computed alongside it. . Worth 1 point.

Part C 3 points

Applies the exact error formula $S-S_n=S\,r^{\,n}$ and argues that with $S>0$ and $0<r<1$ this is always positive. . Worth 2 points. needs an explanation, not just an answer

Concludes that $S_n<S$ for every $n$ , so a partial sum can never reach or overshoot the true sum. . Worth 1 point.
6. A series and a repeating decimal . 11 points. Question 6 of 10.

This question examines two series: one given directly by its terms, and one hidden inside a repeating decimal.
1. Part A.
  Determine whether the series $14 - 21 + 31.5 - 47.25 + \cdots$ converges, stating $r$ and your verdict.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Convert the repeating decimal $0.\overline{63}$ to a fraction in lowest terms by summing the infinite geometric series it represents.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Both $r=0.9$ and $r=-0.9$ satisfy $|r|<1$ . Without computing either sum, decide whether both choices guarantee convergence, and explain how the approach to the limit would look different between the two cases.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$r=-1.5$ ; since $|r|>1$ , the series diverges.

Part B

$0.\overline{63}=\tfrac{7}{11}$ .

Part C

Both converge, since $|0.9|<1$ and $|-0.9|<1$ . With $r=0.9$ the partial sums approach the limit steadily from one side, never crossing it; with $r=-0.9$ they hop back and forth across it, landing closer on each side every time.

Worked solution

Part A

Divide consecutive terms.

$\frac{-21}{14}=-1.5, \qquad \frac{31.5}{-21}=-1.5.$

The ratio holds at $r=-1.5$ , and $|-1.5|=1.5>1$ , so the series diverges.

Part B

Each two-digit block is worth $\tfrac{1}{100}$ of the block before it.

$0.\overline{63} = \frac{63}{100}+\frac{63}{10000}+\cdots, \qquad a=\frac{63}{100},\ r=\frac{1}{100}.$

$S = \frac{63/100}{1-1/100} = \frac{63}{99} = \frac{7}{11}.$

Part C

The convergence test only asks about the size of $r$ , not its sign, so both $r=0.9$ and $r=-0.9$ satisfy $|r|<1$ and both series converge.

The behavior of the partial sums differs, though, and everything traces back to the sign of $r^n$ in the exact error formula.

$S-S_n = S\,r^{\,n}$

With $0<r<1$ , every power $r^n$ is positive, so $S-S_n$ never changes sign: the partial sums approach $S$ from one side only, moving steadily closer without ever crossing it. With $-1<r<0$ , the sign of $r^n$ alternates with the parity of $n$ , so $S-S_n$ alternates sign too: the partial sums overshoot $S$ , then undershoot it, then overshoot again, each hop smaller than the last, closing in on $S$ from both sides at once.

In one line

$14-21+31.5-47.25+\cdots$ has $r=-1.5$ , so $|r|>1$ and it diverges. $0.\overline{63}=\tfrac{7}{11}$ . Both $r=0.9$ and $r=-0.9$ satisfy $|r|<1$ and converge, but a positive ratio approaches the sum from one side while a negative ratio alternates above and below it, closing in from both sides at once.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Finds the ratio by dividing consecutive terms and checks it against a second pair. . Worth 2 points.

Compares $|r|$ with $1$ and reports the correct convergence verdict. . Worth 1 point.

Part B 4 points

Reads the repeating block as a geometric series with first term and ratio both powers of ten. . Worth 2 points.

Applies the sum formula and reduces the result to lowest terms. . Worth 2 points.

Part C 4 points

States that both series converge, since the convergence test depends only on the size of $r$ , not its sign. . Worth 1 point.

Explains that a positive ratio approaches $S$ from one side while a negative ratio alternates above and below $S$ , tying each case to the sign of $r^n$ . . Worth 3 points. needs an explanation, not just an answer
7. How the size of the ratio changes how many terms are enough . 13 points. Question 7 of 10.

A series begins $200 + 40 + 8 + 1.6 + \cdots$ .
1. Part A.
  Find $a$ , $r$ , and $S$ for this series.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Find the smallest whole number of terms $n$ for which $S_n$ is within $0.01$ of $S$ .
  
  Carry your own answer forward Use your own value of $S$ from part A to build the inequality.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  A second series has the same first term $a=200$ and the same tolerance $0.01$ , but a larger ratio, $r=0.5$ . Without solving the new inequality outright, explain whether reaching that tolerance would take more terms or fewer than the number of terms you found in part B, using how the size of $r$ affects how quickly $r^n$ shrinks.
  
  Carry your own answer forward Compare against the number of terms you found in part B; you do not need to solve the new inequality to answer this.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$a=200$ , $r=0.2$ , $S=250$ .

Part B

$n=7$ .

Part C

More terms would be required: a larger ratio shrinks more slowly with each additional term, and the larger ratio also gives a larger sum $S=400$ (versus $S=250$ ), so the second series must reach an even smaller bound. Both effects mean more terms are needed, not fewer.

Worked solution

Part A

Divide consecutive terms: $\tfrac{40}{200}=0.2$ and $\tfrac{8}{40}=0.2$ confirm $r=0.2$ .

$S = \frac{200}{1-0.2} = \frac{200}{0.8} = 250.$

Part B

The exact error after $n$ terms is $S\,r^n$ .

$250(0.2)^n < 0.01 \quad\Longrightarrow\quad n > \frac{\log(0.00004)}{\log(0.2)} \approx 6.29.$

Checking the boundary: $250(0.2)^6=0.016>0.01$ , while $250(0.2)^7=0.0032<0.01$ , so the smallest such $n$ is $7$ .

Part C

Each additional term multiplies the previous power of $r$ by $r$ itself, and a ratio closer to $1$ shrinks that power more slowly than a ratio closer to $0$ does.

$r^{\,n+1} = r \cdot r^{\,n},$

so a bigger $r$ (closer to $1$ ) shrinks $r^n$ toward $0$ more slowly with each extra factor. Since $0.5$ is larger than $0.2$ , the sequence $(0.5)^1,(0.5)^2,(0.5)^3,\ldots$ falls toward $0$ more slowly than $(0.2)^1,(0.2)^2,(0.2)^3,\ldots$ does.

For each series the exact requirement is $r^n < \tfrac{0.01}{S}$ , and that bound is not the same for both series: with $r=0.5$ , $S=\tfrac{200}{1-0.5}=400$ , a larger sum than the $S=250$ found in part A, so the second series must reach an even smaller bound, $\tfrac{0.01}{400}$ , than the first series' $\tfrac{0.01}{250}$ . Both effects point the same way, a more slowly shrinking $r^n$ and a tighter bound to reach, so the series with $r=0.5$ needs strictly MORE terms than the number found in part B, not fewer.

In one line

For $200+40+8+1.6+\cdots$ : $a=200$ , $r=0.2$ , $S=250$ , and the smallest $n$ with $S_n$ within $0.01$ of $S$ is $n=7$ . A series with the same $a$ and tolerance but a larger ratio $r=0.5$ would need MORE terms than that, because a ratio closer to $1$ makes $r^n$ shrink more slowly with each added term, and because the larger ratio also gives a larger sum $S=400$ , requiring an even smaller bound to be reached.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Finds the ratio and checks it against a second pair of terms. . Worth 1 point.

Confirms $|r|<1$ and applies the sum formula correctly to get $S=250$ . . Worth 2 points.

Reports $S$ as the number the partial sums close in on, not as a term of the series. . Worth 1 point.

Part B 5 points

Sets up the correct inequality from the exact error formula $S-S_n=S\,r^n$ and the tolerance $0.01$ . . Worth 2 points.

Solves with logarithms and rounds up to the correct smallest whole number of terms, checking the boundary. . Worth 2 points.

Reports $n$ as a whole number of terms, not the raw decimal solution to the inequality. . Worth 1 point.

Part C 4 points

States the correct direction: a larger ratio requires more terms, not fewer. . Worth 1 point.

Justifies the direction using how quickly $r^n$ shrinks as a function of the size of $r$ , rather than by solving the new inequality. . Worth 3 points. needs an explanation, not just an answer
8. Which terms of an expansion flip sign . 11 points. Question 8 of 10.

This question expands a binomial difference and checks the result before asking which of its terms are negative.
1. Part A.
  Expand $(y-4)^4$ completely.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Check your expansion by substituting $y=1$ into both the original expression and your expanded form, and confirming the two agree.
  
  Carry your own answer forward Substitute $y=1$ into your own expansion from part A, even if it was not the expected one.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Which terms of your expansion are negative? Explain, in general, why exactly those terms come out negative and not some other subset.
  
  Carry your own answer forward Identify the negative terms from your own expansion in part A.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$(y-4)^4=y^4-16y^3+96y^2-256y+256$ .

Part B

Substituting $y=1$ gives $81$ on both sides, confirming the expansion.

Part C

The odd- $k$ terms, $-16y^3$ and $-256y$ , are negative; $(-4)^k$ is negative exactly when $k$ is odd, so those are the only terms whose sign flips relative to $(y+4)^4$ .

Worked solution

Part A

Apply the theorem with $A=y$ , $B=-4$ , $n=4$ , using the coefficients $1,4,6,4,1$ .

$(y-4)^4 = y^4+4y^3(-4)+6y^2(-4)^2+4y(-4)^3+(-4)^4 = y^4-16y^3+96y^2-256y+256.$

Part B

The original expression gives $(1-4)^4=(-3)^4=81$ . The expanded form gives

$1-16+96-256+256 = 81,$

which matches.

Part C

The negative terms are $-16y^3$ and $-256y$ , at $k=1$ and $k=3$ .

Every term carries a factor of $(-4)^k$ , and that factor splits into a sign and a size.

$(-4)^k = (-1)^k 4^k.$

Since $(-1)^k$ is $-1$ exactly when $k$ is odd and $+1$ when $k$ is even, only the odd- $k$ terms, $k=1$ and $k=3$ here, pick up the extra minus sign. The even- $k$ terms, $k=0,2,4$ , are unaffected by the change from $+4$ to $-4$ .

In one line

$(y-4)^4=y^4-16y^3+96y^2-256y+256$ , checked at $y=1$ where both sides equal $81$ . The negative terms are $-16y^3$ and $-256y$ , at odd $k$ , since $(-4)^k=(-1)^k4^k$ is negative exactly when $k$ is odd.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Writes the general term with the row-4 coefficients before combining anything. . Worth 1 point.

Evaluates every power of $-4$ and every coefficient correctly, arriving at the complete expansion. . Worth 3 points.

Part B 3 points

Evaluates the original expression at $y=1$ to get $81$ . . Worth 1 point.

Evaluates the expanded form at $y=1$ and confirms it agrees with the original. . Worth 2 points.

Part C 4 points

Correctly identifies the two negative terms from the expansion. . Worth 1 point.

Explains the split using the parity of $k$ , tying it to $(-1)^k$ . . Worth 3 points. needs an explanation, not just an answer
9. Checking one coefficient two ways . 12 points. Question 9 of 10.

$(2a-b)^6$ is expanded two different ways, and the two are then compared.
1. Part A.
  Expand $(2a-b)^6$ completely.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Without redoing the expansion, use the general term formula to verify the coefficient of the $a^2b^4$ term you found in part A.
  
  Carry your own answer forward Compare your result here to the $a^2b^4$ term in your own expansion from part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Suppose only the coefficient of $a^2b^4$ had been needed, with nothing else from the expansion. Explain why the general term formula from part B is the more efficient tool for that specific task, even though the full expansion from part A also contains the answer.
  
  Carry your own answer forward Compare the work in part B against the work in part A.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 4 points
The answer

Part A

$(2a-b)^6=64a^6-192a^5b+240a^4b^2-160a^3b^3+60a^2b^4-12ab^5+b^6$ .

Part B

$T_4=\binom{6}{4}(2a)^2(-b)^4=60a^2b^4$ , matching part A.

Part C

The general term is faster here because it produces the one requested coefficient directly from a single substitution, while the full expansion computes six other coefficients that this task never needed.

Worked solution

Part A

Apply the theorem with $A=2a$ , $B=-b$ , $n=6$ , using the coefficients $1,6,15,20,15,6,1$ .

$(2a-b)^6 = \sum_{k=0}^{6}\binom{6}{k}(2a)^{6-k}(-b)^k = 64a^6-192a^5b+240a^4b^2-160a^3b^3+60a^2b^4-12ab^5+b^6.$

Part B

The $a^2b^4$ term needs $6-k=2$ , so $k=4$ .

$T_4 = \binom{6}{4}(2a)^2(-b)^4 = 15\cdot4a^2\cdot b^4 = 60a^2b^4,$

which matches the $a^2b^4$ term found in part A.

Part C

The full expansion in part A produces all seven terms of $(2a-b)^6$ at once, which is exactly what is needed when every term matters, but it computes six coefficients beyond the one asked for here. The general term formula in part B skips straight to the single term wanted: solving one equation for $k$ and substituting once produces the requested coefficient directly, with no other term ever computed.

$6-k=2 \;\Longrightarrow\; k=4 \;\Longrightarrow\; T_4 = 60a^2b^4,$

with no $T_0,T_1,T_2,T_3,T_5,T_6$ ever computed along the way. When only one coefficient is needed, the general term does the same job as the full expansion with far less work; the full expansion earns its cost only when the whole polynomial, not one piece of it, is actually wanted.

In one line

$(2a-b)^6=64a^6-192a^5b+240a^4b^2-160a^3b^3+60a^2b^4-12ab^5+b^6$ . The general term at $k=4$ gives $T_4=60a^2b^4$ , matching part A. When only one coefficient is needed, the general term formula is the more efficient tool, since it produces that single term directly instead of computing the six others the full expansion also finds.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Writes the general term $\binom{6}{k}(2a)^{6-k}(-b)^k$ for each $k$ before combining anything. . Worth 1 point.

Evaluates every power of $2a$ , every power of $-b$ , and every coefficient correctly, arriving at all seven terms. . Worth 3 points.

Part B 4 points

Sets $6-k=2$ and solves for the correct $k=4$ . . Worth 1 point.

Evaluates the general term at $k=4$ and confirms it matches the corresponding term from part A. . Worth 2 points.

States explicitly that the two methods agree, not just that a number was produced. . Worth 1 point.

Part C 4 points

Explains that the general term formula produces only the requested coefficient, without computing the other terms the full expansion produces. . Worth 3 points. needs an explanation, not just an answer

Notes that the full expansion is still the right tool when every term of the polynomial, not just one, is actually needed. . Worth 1 point.
10. Solving for an unknown row from two of its coefficients . 11 points. Question 10 of 10.

This question solves for an unknown row number $n$ from a relationship between two of its coefficients.
1. Part A.
  Using $\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$ , write $\binom{n}{2}$ and $\binom{n}{3}$ each as a simplified expression in $n$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Solve $\binom{n}{3}=4\binom{n}{2}$ for the whole number $n\ge3$ , and check your value.
  
  Carry your own answer forward Use the two expressions you found in part A to set up the equation.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  If the equation from part B is cleared of fractions without first dividing by $n(n-1)$ , the result factors completely into three linear factors in $n$ , and one of them reproduces the value you found in part B. State the other two whole-number roots of the cleared equation. Then decide whether they represent valid row numbers here, and explain your reasoning.
  
  Carry your own answer forward The cleared equation follows from the one you set up in part B.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$\binom{n}{2}=\tfrac{n(n-1)}{2}$ ; $\binom{n}{3}=\tfrac{n(n-1)(n-2)}{6}$ .

Part B

$n=14$ ; check: $\binom{14}{3}=364=4\binom{14}{2}=4(91)$ .

Part C

The other two roots are $n=0$ and $n=1$ , and neither is valid: $\binom{n}{3}$ requires choosing $3$ items from a set of $n$ , which only makes sense when $n\ge3$ , so those roots satisfy the cleared algebraic equation without corresponding to a genuine value of the original binomial coefficients.

Worked solution

Part A

Cancel the shared factorial tail in each closed form.

$\binom{n}{2}=\frac{n!}{2!(n-2)!}=\frac{n(n-1)(n-2)!}{2(n-2)!}=\frac{n(n-1)}{2}, \qquad \binom{n}{3}=\frac{n!}{3!(n-3)!}=\frac{n(n-1)(n-2)}{6}.$

Part B

Substitute the expressions from part A.

$\frac{n(n-1)(n-2)}{6} = 4\cdot\frac{n(n-1)}{2} = 2n(n-1).$

Since $n\ge3$ makes $n(n-1)\ne0$ , divide both sides by it.

$\frac{n-2}{6}=2 \quad\Longrightarrow\quad n-2=12 \quad\Longrightarrow\quad n=14.$

Check: $\binom{14}{2}=91$ and $\binom{14}{3}=364=4(91)$ .

Part C

Setting each factor of $n(n-1)(n-14)=0$ to zero gives three candidate roots.

$n=0, \qquad n=1, \qquad n=14.$

But $n=0$ and $n=1$ are not valid for the ORIGINAL equation. The closed forms from part A, $\binom{n}{2}=\tfrac{n(n-1)}{2}$ and $\binom{n}{3}=\tfrac{n(n-1)(n-2)}{6}$ , are algebraic expressions that can be evaluated at any $n$ , but $\binom{n}{k}$ itself, as a count of ways to choose $k$ items from $n$ , is only defined when $0\le k\le n$ . Choosing $3$ items requires a set of at least $3$ to choose from, so $\binom{n}{3}$ has no meaning at all when $n=0$ or $n=1$ , even though its closed-form expression can still be evaluated there.

Clearing fractions turned the original relationship between two binomial coefficients into a purely algebraic polynomial equation, and that wider equation picked up two extra roots that the closed-form expressions satisfy but that the original counting problem never could. That is exactly why the problem restricts attention to $n\ge3$ in the first place: it is the domain where $\binom{n}{2}$ and $\binom{n}{3}$ are both genuinely defined, and $n=14$ is the only root of the cleared equation that lands inside it.

In one line

$\binom{n}{2}=\tfrac{n(n-1)}{2}$ and $\binom{n}{3}=\tfrac{n(n-1)(n-2)}{6}$ ; solving $\binom{n}{3}=4\binom{n}{2}$ gives $n=14$ (checked: $364=4(91)$ ). The cleared equation $n(n-1)(n-14)=0$ also has roots $n=0$ and $n=1$ , but $\binom{n}{3}$ is only defined when $n\ge3$ , so those roots satisfy the algebraic equation without describing a genuine row and are excluded by the problem's restriction.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Simplifies $\binom{n}{2}$ to a fully reduced expression in $n$ . . Worth 1 point.

Simplifies $\binom{n}{3}$ to a fully reduced expression in $n$ , showing the cancellation. . Worth 2 points.

Part B 4 points

Sets up the equation matching $\binom{n}{3}$ to $4\binom{n}{2}$ using the expressions from part A. . Worth 1 point.

Solves for $n=14$ by a valid method. . Worth 2 points.

Checks the answer by evaluating both sides of the original equation at $n=14$ . . Worth 1 point.

Part C 4 points

States the two other roots correctly, $n=0$ and $n=1$ . . Worth 1 point.

Explains that $\binom{n}{3}$ requires $n\ge3$ to be defined at all, so $n=0$ and $n=1$ satisfy the cleared algebraic equation without corresponding to a genuine value of the original binomial coefficients, and connects this to the problem's restriction $n\ge3$ . . Worth 3 points. needs an explanation, not just an answer

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. A sum, relabeled and shifted 12 points

Part A 4

Identifies $j^2$ as the summand, matching the pattern of the given terms. 2
Sets the lower limit at $4$ and the upper limit at $10$ , matching the first and last given terms. 2

Part B 4

Reports the term count, correctly computed from both limits of the sum. 1
Evaluates the summand at every index from $4$ to $10$ and adds all seven results correctly. 2
Reports the term count and the sum as two separate, clearly labeled quantities. 1

Part C 4

Correctly substitutes $j=p+4$ into the summand and both limits, and confirms the rewritten sum matches the total from part B. 2
Explains that the substitution merely relabels the same list of values in the same order, so the sum's value cannot change. 2

2. A shrinking sequence and how soon it drops below a bound 12 points

Part A 4

Finds the ratio by dividing consecutive terms and checks it against a second pair. 1
Writes the explicit formula and uses the exponent $7$ , not $8$ , when substituting $n=8$ . 2
Reduces $a_8$ to lowest terms. 1

Part B 4

Selects the finite-sum formula with the correct $a$ , $r$ , and $n$ . 1
Carries the fraction arithmetic through correctly to $S_8=\tfrac{6560}{243}$ . 2
Reports the result reduced to lowest terms. 1

Part C 4

Sets up and solves the correct inequality with logarithms, handling the reversal from dividing by a negative logarithm. 2
Explains, from $0<r<1$ , why the sequence strictly decreases and so cannot climb back above the bound once it has dropped below it. 2

3. A rule that reaches back one step, and the shortcut hiding inside it 11 points

Part A 3

Builds $k_2$ , $k_3$ , $k_4$ in order from the single seed $k_1$ . 2
Confirms the common difference is constant across the computed terms and states $d=6$ . 1

Part B 4

Substitutes correctly and simplifies to $k_n=6n+5$ . 2
Evaluates the simplified formula at $n=15$ to get $k_{15}=95$ . 1
Reports the result as the value of the 15th term, not the position number itself. 1

Part C 4

States that a recursive rule and an explicit formula are two different ways of describing the same sequence, not two competing sequences. 2
Supports the claim by checking that $k_n=6n+5$ reproduces the seed and satisfies the recursive step, or gives an equivalent concrete check. 2

4. Which two partial sums actually isolate a range 13 points

Part A 4

Uses $19-1=18$ steps in the explicit formula, not $19$ . 1
Computes $-8+18(5)=82$ correctly. 2
Reports the result as the value of the 19th term, not the position number itself. 1

Part B 5

Counts the range from $a_9$ through $a_{19}$ inclusively as $11$ terms and finds $a_9=32$ . 2
Applies the sum formula to the $11$ -term range correctly to get $627$ . 2
Reports the result as the total for the stated range, not as $S_{19}$ or any other full running total. 1

Part C 4

Identifies that $S_9$ already includes $a_9$ , so subtracting it removes a term that belongs in the range. 2
States that $S_{19}-S_8$ is the correct pair, and confirms it reproduces the total from part B. 2

5. A finite total, and how close it already sits to the infinite one 12 points

Part A 4

Selects the finite-sum formula with the correct $a$ , $r$ , $n$ . 1
Carries the fraction arithmetic through correctly to $S_6=\tfrac{4095}{64}$ . 2
Reports the result reduced to lowest terms. 1

Part B 5

Confirms $|r|<1$ and applies the infinite sum formula to get $S=64$ . 2
Computes the exact gap $S-S_6=\tfrac{1}{64}$ , by the error formula or by direct subtraction. 2
Reports $S$ as the number the partial sums close in on, distinct from the gap $S-S_6$ computed alongside it. 1

Part C 3

Applies the exact error formula $S-S_n=S\,r^{\,n}$ and argues that with $S>0$ and $0<r<1$ this is always positive. 2
Concludes that $S_n<S$ for every $n$ , so a partial sum can never reach or overshoot the true sum. 1

6. A series and a repeating decimal 11 points

Part A 3

Finds the ratio by dividing consecutive terms and checks it against a second pair. 2
Compares $|r|$ with $1$ and reports the correct convergence verdict. 1

Part B 4

Reads the repeating block as a geometric series with first term and ratio both powers of ten. 2
Applies the sum formula and reduces the result to lowest terms. 2

Part C 4

States that both series converge, since the convergence test depends only on the size of $r$ , not its sign. 1
Explains that a positive ratio approaches $S$ from one side while a negative ratio alternates above and below $S$ , tying each case to the sign of $r^n$ . 3

7. How the size of the ratio changes how many terms are enough 13 points

Part A 4

Finds the ratio and checks it against a second pair of terms. 1
Confirms $|r|<1$ and applies the sum formula correctly to get $S=250$ . 2
Reports $S$ as the number the partial sums close in on, not as a term of the series. 1

Part B 5

Sets up the correct inequality from the exact error formula $S-S_n=S\,r^n$ and the tolerance $0.01$ . 2
Solves with logarithms and rounds up to the correct smallest whole number of terms, checking the boundary. 2
Reports $n$ as a whole number of terms, not the raw decimal solution to the inequality. 1

Part C 4

States the correct direction: a larger ratio requires more terms, not fewer. 1
Justifies the direction using how quickly $r^n$ shrinks as a function of the size of $r$ , rather than by solving the new inequality. 3

8. Which terms of an expansion flip sign 11 points

Part A 4

Writes the general term with the row-4 coefficients before combining anything. 1
Evaluates every power of $-4$ and every coefficient correctly, arriving at the complete expansion. 3

Part B 3

Evaluates the original expression at $y=1$ to get $81$ . 1
Evaluates the expanded form at $y=1$ and confirms it agrees with the original. 2

Part C 4

Correctly identifies the two negative terms from the expansion. 1
Explains the split using the parity of $k$ , tying it to $(-1)^k$ . 3

9. Checking one coefficient two ways 12 points

Part A 4

Writes the general term $\binom{6}{k}(2a)^{6-k}(-b)^k$ for each $k$ before combining anything. 1
Evaluates every power of $2a$ , every power of $-b$ , and every coefficient correctly, arriving at all seven terms. 3

Part B 4

Sets $6-k=2$ and solves for the correct $k=4$ . 1
Evaluates the general term at $k=4$ and confirms it matches the corresponding term from part A. 2
States explicitly that the two methods agree, not just that a number was produced. 1

Part C 4

Explains that the general term formula produces only the requested coefficient, without computing the other terms the full expansion produces. 3
Notes that the full expansion is still the right tool when every term of the polynomial, not just one, is actually needed. 1

10. Solving for an unknown row from two of its coefficients 11 points

Part A 3

Simplifies $\binom{n}{2}$ to a fully reduced expression in $n$ . 1
Simplifies $\binom{n}{3}$ to a fully reduced expression in $n$ , showing the cancellation. 2

Part B 4

Sets up the equation matching $\binom{n}{3}$ to $4\binom{n}{2}$ using the expressions from part A. 1
Solves for $n=14$ by a valid method. 2
Checks the answer by evaluating both sides of the original equation at $n=14$ . 1

Part C 4

States the two other roots correctly, $n=0$ and $n=1$ . 1
Explains that $\binom{n}{3}$ requires $n\ge3$ to be defined at all, so $n=0$ and $n=1$ satisfy the cleared algebraic equation without corresponding to a genuine value of the original binomial coefficients, and connects this to the problem's restriction $n\ge3$ . 3

Multiple choice

Free response