Sequences and Series: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

A sequence is defined by the explicit rule $b_n = 2n^2 - n + 3$ . What is $b_5$ ?

Answer choices for question 1 A $98$ B $58$ C $48$ D $45$
2

An arithmetic sequence has $a_1 = 7$ and $d = 5$ . What is $a_9$ ?

Answer choices for question 2 A $47$ B $52$ C $42$ D $45$
3

A geometric sequence has $a_1 = 4$ and $r = 3$ . What is $a_6$ ?

Answer choices for question 3 A $2916$ B $19$ C $60$ D $972$
4

Evaluate $\displaystyle\sum_{k=3}^{7} (2k+1)$ .

Answer choices for question 4 A $40$ B $55$ C $72$ D $60$
5

A sequence is given by $g_1 = 2$ and $g_n = g_{n-1} + 4$ for $n \ge 2$ . What is $g_1+g_2+g_3+g_4$ ?

Answer choices for question 5 A $30$ B $48$ C $32$ D $50$
6

Find the sum of the series $12 - 4 + \tfrac43 - \tfrac49 + \cdots$ .

Answer choices for question 6 A $9$ B $18$ C The series diverges, since inverting the ratio gives $|r|=3>1$ . D $16$
7

Advanced. This question goes beyond core Algebra II. It is not required by the course.

What is the coefficient of $x^2$ in the expansion of $(x+2)^4$ ?

Answer choices for question 7 A $12$ B $24$ C $16$ D $4$
8

An arithmetic sequence has $a_3 = 17$ and $a_{10} = 59$ . What is $d$ ?

Answer choices for question 8 A $-6$ B $4.2$ C $5.25$ D $6$
9

Evaluate $\displaystyle\sum_{k=4}^{9} (3k-2)$ .

Answer choices for question 9 A $105$ B $87.5$ C $114$ D $117$
10

Advanced. This question goes beyond core Algebra II. It is not required by the course.

What is the coefficient of $x^4y^3$ in the expansion of $(x+3y)^7$ ?

Answer choices for question 10 A $105$ B $2835$ C $945$ D $315$
11

A geometric sequence has all positive terms, with $a_2 = 4$ and $a_6 = 324$ . What is $r$ ?

Answer choices for question 11 A $-3$ B $3$ C $81$ D $\dfrac13$
12

A sequence begins $2, 12, 72, 432, \ldots$ . What is $a_7$ ?

Answer choices for question 12 A $62$ B $559872$ C $\dfrac{1}{23328}$ D $93312$
13

A sequence is given by $h_1 = 5$ , $h_2 = 8$ , and $h_n = h_{n-1} - h_{n-2}$ for $n \ge 3$ . What is $h_6$ ?

Answer choices for question 13 A $-3$ B $25$ C $3$ D $-8$
14

What happens to the partial sums of the series $11 - 11 + 11 - 11 + \cdots$ ?

Answer choices for question 14 A The partial sums grow without bound, so the series diverges by running away. B The series converges to $0$ , since the positive and negative terms cancel in pairs. C The partial sums alternate forever between $11$ and $0$ without ever settling on one value, so the series diverges. D The series converges to $\dfrac{11}{2}$ by $S=\dfrac{a}{1-r}$ , since $a=11$ and $r=-1$ .
15

Find the sum of the series $20 + 20\left(\tfrac34\right) + 20\left(\tfrac34\right)^2 + \cdots$ .

Answer choices for question 15 A $\dfrac{80}{7}$ B $80$ C $\dfrac{80}{3}$ D $46.25$
16

Advanced. This question goes beyond core Algebra II. It is not required by the course.

What is the coefficient of $x^3$ in the expansion of $(2x-1)^5$ ?

Answer choices for question 16 A $20$ B $-80$ C $-40$ D $80$
17

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Which of the following is the coefficient of the $x^3$ term in the expansion of $\left(x^2+\dfrac{3}{x}\right)^5$ ?

Answer choices for question 17 A No such term exists, since $10-3k=3$ gives $k=\tfrac{7}{3}$ , not a whole number. B $270$ C $90$ D $405$
18

A series has $a=180$ and $r=0.4$ . What is the smallest whole number of terms $n$ for which $S_n$ is within $1$ of the series' sum?

Answer choices for question 18 A $6$ B $8$ C $7$ D $2$
19

A sequence's partial sums satisfy $S_n = 4n^2 + 7n$ for every positive integer $n$ . What is its common difference $d$ ?

Answer choices for question 19 A $4$ B $8$ C $7$ D $11$
20

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Row $7$ of Pascal's triangle is built from row $6$ , $1,6,15,20,15,6,1$ , using Pascal's rule. What is the coefficient of $x^4y^3$ in the expansion of $(x-2y)^7$ ?

Answer choices for question 20 A $280$ B $-70$ C $-160$ D $-280$

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Where you stand

Construct and evaluate a sequence's explicit rule, and construct and evaluate its recursive rule, including that a rule looking back $k$ steps needs $k$ seed values before it defines anything. Review: Sequences and Notation
Evaluate a sum given in sigma notation, write a sum in sigma notation, count its terms as $p-m+1$ with both limits included, and treat the index of summation as a renameable dummy variable. Review: Sequences and Notation
Apply $a_n=a_1+(n-1)d$ and recover $d$ from two terms with $d=\dfrac{a_m-a_k}{m-k}$ ; apply the sum formulas $S_n=\dfrac{n}{2}(a_1+a_n)=\dfrac{n}{2}\bigl(2a_1+(n-1)d\bigr)$ ; and use that a sequence is arithmetic exactly when its partial sum $S_n$ is quadratic in $n$ with no constant term. Review: Arithmetic Sequences and Series
Apply $a_n=a_1r^{\,n-1}$ and recover $r$ from two terms; sum a finite geometric series with $S_n=\dfrac{a(1-r^{\,n})}{1-r}$ , switching to $S_n=na$ at $r=1$ ; read a fixed percentage change as $r$ ; and find how many terms a threshold requires using logarithms. Review: Geometric Sequences and Series
Decide whether an infinite geometric series converges using $|r|<1$ , compute its sum with $S=\dfrac{a}{1-r}$ , distinguish partial sums that pass every bound from partial sums that stay bounded but never settle, and read a repeating decimal as such a series. Review: Infinite Geometric Series
Use the exact error $S-S_n=S\,r^{\,n}$ and logarithms to find how many terms of a convergent series are needed to approximate its sum within a stated tolerance. Review: Infinite Geometric Series
Expand $(x+y)^n$ and $(x-y)^n$ using the coefficients of Pascal's triangle, tracking the alternating sign $(-1)^k$ a difference introduces, and build an unfamiliar row from the row above it with Pascal's rule. Review: The Binomial Theorem
Use the general term $T_k=\binom{n}{k}A^{\,n-k}B^{\,k}$ to extract one term or coefficient of an expansion by solving for $k$ from a target exponent, recognizing when no whole-number $k$ exists, and use $\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$ to solve for an unknown row number $n$ . Review: The Binomial Theorem

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Read backward

A fourteen-term arithmetic sequence is read backward. In that order, its first term is $17$ and its common difference is $-\frac52$ . Find the first term and the sum of the sequence in its original order.
1. Hint 1
  Reading a list backward reverses the sign of its common difference and swaps its two ends.
2. Hint 2
  The last term of the backward reading is the first term of the original.
Answer

First term $-\frac{31}2$ ; sum $\frac{21}2$ .

Full solution

Backward, the fourteenth term is $17+13(-\frac52)=17-\frac{65}2$ , which is $-\frac{31}2$ .
That is the original first term, and the original common difference is $\frac52$ .

The sum does not depend on the order:
$S_{14}=\frac{14}2\left(17-\frac{31}2\right)$
which is
$7\cdot\frac32=\frac{21}2.$

Answer

First term $-\frac{31}2$ ; sum $\frac{21}2$ .

Key idea

Reversing an arithmetic list negates its common difference and keeps its sum.
Problem 1: Mark complete
Problem 2 A growing record

An increasing geometric sequence has positive terms, with $a_2=16$ and $a_1+a_3=68$ . What is the first index $n$ for which $a_n>1000$ , and what is the sum of all terms from $a_1$ through that term?
1. Hint 1
  Write the first and third terms through the second term and the ratio.
2. Hint 2
  After recovering the ratio and first term, take logarithms of the threshold inequality and check neighboring indices; then total exactly that many terms.
Answer

$n=5$ ; the sum through $a_5$ is $1364$ .

Full solution

With ratio $r>0$ , $a_1=\frac{16}r$ and $a_3=16r$ , so
$\frac{16}r+16r=68.$
Multiplying by $\frac r4$ gives $4r^2-17r+4=0$ , that is,
$(4r-1)(r-4)=0.$
The sequence increases, so $r=4$ rather than $\frac14$ , and $a_1=4$ .
Thus $a_n=4^n$ .

The threshold $4^n>1000$ gives
$n>\frac{\ln1000}{\ln4}$
The right side is approximately $4.98$ , so the first possible integer is $n=5$ .
Checking the boundary, $a_4=256$ does not exceed $1000$ , whereas $a_5=1024$ does.

Their total is
$S_5=\frac{4(4^5-1)}{4-1}$
which is
$\frac{4\cdot1023}3=1364.$
As a check,
$4+16+64+256+1024=1364.$

Answer

$n=5$ ; the sum through $a_5$ is $1364$ .

Key idea

A threshold index determines how many terms belong in the corresponding finite geometric total.
Problem 2: Mark complete
Problem 3 Three symbols

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Expand $((x+1)+y)^2$ fully as a polynomial in $x$ and $y$ .
1. Hint 1
  A complete expression can serve as one term of a binomial.
2. Hint 2
  Expand the outside square, then collect the terms from the square and product involving $x+1$ .
Answer

$x^2+2xy+y^2+2x+2y+1$ .

Full solution

With the two terms $x+1$ and $y$ , the outside square is
$(x+1)^2+2(x+1)y+y^2$
Expanding the remaining square and product gives $x^2+2x+1+2xy+2y+y^2$ .

Collecting and arranging the terms gives $x^2+2xy+y^2+2x+2y+1$ .
Setting $x=y=1$ gives $9$ in both forms, a consistency check.

Answer

$x^2+2xy+y^2+2x+2y+1$ .

Key idea

Binomial expansion can be applied repeatedly when one of its terms is itself a sum.
Problem 3: Mark complete
Problem 4 A two-track record

A sequence is defined for integers $n\ge2$ by $a_2=8$ , $a_3=1$ , and $a_n=a_{n-2}+4$ for $n\ge4$ . Find an explicit rule. Write and evaluate $a_5+\cdots+a_{10}$ in sigma notation, state its term count, and rewrite that sigma expression with index $j$ . Explain why both given starting values are needed.
1. Hint 1
  The recursion advances the even and odd indices separately.
2. Hint 2
  Find a separate explicit rule for each parity, then check it against its seed and the recursion; both endpoints of the requested sum count.
Answer

$a_n=2n+4$ for even $n\ge2$ and $a_n=2n-5$ for odd $n\ge3$ ; $\sum_{k=5}^{10}a_k=87$ ; $6$ terms; equivalently $\sum_{j=5}^{10}a_j$ . $a_2$ seeds the even chain and $a_3$ the odd chain.

Full solution

Even indices run $a_2=8$ , $a_4=12$ , $a_6=16$ , and so on, rising by $4$ every two indices, so $a_n=2n+4$ for even $n\ge2$ .
Odd indices run $a_3=1$ , $a_5=5$ , $a_7=9$ , so $a_n=2n-5$ for odd $n\ge3$ .
Each rule matches its seed and rises by $4$ from $n-2$ to $n$ .

There are $10-5+1=6$ terms: $5,16,9,20,13,24$ , whose sum is $87$ .
Thus $\sum_{k=5}^{10}a_k=87$ , and replacing the dummy index $k$ by $j$ leaves the limits and the value unchanged.

The recursion links each term only to the term two places back, so the even and odd indices form separate chains.
The seed $a_2$ fixes every even-indexed term and $a_3$ every odd-indexed one; neither can be found from the other.

Answer

$a_n=2n+4$ for even $n\ge2$ and $a_n=2n-5$ for odd $n\ge3$ ; $\sum_{k=5}^{10}a_k=87$ ; $6$ terms; equivalently $\sum_{j=5}^{10}a_j$ . $a_2$ seeds the even chain and $a_3$ the odd chain.

Key idea

A recursion that advances separate index chains needs a seed for each chain, while summation indices are only names.
Problem 4: Mark complete
Problem 5 Two supply plans

Plan A supplies amounts in an arithmetic sequence: the second day supplies $22$ liters and the fifth supplies $28$ liters. Plan B supplies $24$ liters on day one, with a fixed percentage change of zero each day. Which plan supplies more over the first four days, and by how many liters? State the daily difference for A and the daily ratio for B.
1. Hint 1
  The two known days of A are three steps apart.
2. Hint 2
  A zero percentage change leaves the amount unchanged, so its geometric ratio needs the constant-sequence sum case.
Answer

Plan B supplies $4$ liters more; A totals $92$ liters with $d=2$ ; B totals $96$ liters with $r=1$ .

Full solution

For A, the daily difference is $(28-22)/(5-2)=2$ liters.
Its first four amounts are $20,22,24,26$ , giving
$S_4=\frac42(20+26)$
Thus A supplies $92$ liters.

For B the daily multiplier is $1+0/100=1$ .
The quotient formula with denominator $1-r$ would be undefined, so use the constant sum: four lots of $24$ give $96$ liters.
B supplies $96-92=4$ liters more.

Answer

Plan B supplies $4$ liters more; A totals $92$ liters with $d=2$ ; B totals $96$ liters with $r=1$ .

Key idea

A comparison of totals requires the sum formula appropriate to each pattern, including the constant geometric case.
Problem 5: Mark complete
Problem 6 A four-digit repeating block

The decimal $d=0.12341234\ldots$ repeats the four-digit block $1234$ . A student claims $d=\frac{12}{99}+\frac{34}{9900}$ by treating the two two-digit blocks separately. Is the claim correct? Express $d$ as a convergent geometric series and an exact fraction, and identify the error in the proposed block spacing.
1. Hint 1
  The complete repeated block returns every four decimal places.
2. Hint 2
  Use the first four-digit block as the first term and the shift to the next copy as the ratio.
Answer

No; $d=\sum_{k=0}^{\infty}\frac{1234}{10000}(\frac1{10000})^k=\frac{1234}{9999}$ . The error: each two-digit block was repeated every two places instead of every four.

Full solution

The first block contributes $1234/10000$ , and each later copy is $1/10000$ times the preceding one.
Since this ratio has magnitude below one, the series converges.
Its sum is
$d=\frac{1234}{9999}$

The proposed fractions repeat each two-digit block every two places.
In fact $12/99=0.121212\ldots$ and $34/9900=0.00343434\ldots$ , whose sum begins $0.124646\ldots$ .
The intended copies of each block are four places apart, so the proposed spacing is wrong.

Answer

No; $d=\sum_{k=0}^{\infty}\frac{1234}{10000}(\frac1{10000})^k=\frac{1234}{9999}$ . The error: each two-digit block was repeated every two places instead of every four.

Key idea

When a block of b digits repeats, each copy contributes one over ten to the b times the preceding copy's contribution.
Problem 6: Mark complete
Problem 7 Two bounds on a sum

Let $S$ be the sum of $4-3+\frac94-\frac{27}{16}+\cdots$ , and let $S_k$ be its first $k$ terms. Find $S$ . Find the smallest positive odd integer $m$ and the smallest positive even integer $n$ such that $S_n<S<S_m$ and each bound differs from $S$ by less than $0.0001$ . Justify why each smaller term count of the same parity fails.
1. Hint 1
  The sign of the ratio determines which partial sums lie above or below the infinite sum.
2. Hint 2
  Use $|S-S_k|=|S||r|^k$ and logarithms to find an index bound, then enforce the required odd and even choices.
Answer

$S=\frac{16}{7}$ ; the upper bound uses $m=35$ , and the lower bound uses $n=36$ .

Full solution

The ratio is $-3/4$ , with magnitude below one, so the series converges and $S=4/(1+3/4)$ , which is $16/7$ .
Its signed remainder is
$S-S_k=\frac{16}{7}(-\tfrac34)^k$
For odd $k$ this is negative, so the partial sum lies above $S$ ; for even $k$ it is positive, so the partial sum lies below $S$ .

The required absolute error is
$(16/7)(3/4)^k<0.0001.$
Taking logarithms and dividing by the negative $\ln(3/4)$ gives
$k>\frac{\ln(7/160000)}{\ln(3/4)}$
The right side is approximately $34.89$ .
Thus the least qualifying odd count is $35$ and the least qualifying even count is $36$ .

At $k=35$ and $k=36$ , the absolute errors are approximately $0.0000969$ and $0.0000726$ .
The preceding odd count $33$ has error approximately $0.0001722$ , and the preceding even count $34$ has error approximately $0.0001292$ .
Both fail, and all earlier errors are larger because $3/4<1$ .

Answer

$S=\frac{16}{7}$ ; the upper bound uses $m=35$ , and the lower bound uses $n=36$ .

Key idea

With a nonzero first term and a ratio between $-1$ and $0$ , the partial sums alternate above and below the sum, so the required term counts depend on the remainder's magnitude and on whether the term count is odd or even.
Problem 7: Mark complete
Problem 8 A check that misses

Advanced. This question goes beyond core Algebra II. It is not required by the course.

A student expands $(3u-4v)^3$ and assigns the coefficients $27,-107,143,-64$ to $u^3,u^2v,uv^2,v^3$ in that order. At $u=v=1$ , both the proposal and the original give $-1$ . Does this check establish the expansion? Give the correct expansion and identify the incorrect coefficients.
1. Hint 1
  One substitution checks a single number built from all the coefficients at once.
2. Hint 2
  Expand with row 3 of Pascal's triangle and the powers of $3u$ and of $-4v$ .
Answer

No; $(3u-4v)^3=27u^3-108u^2v+144uv^2-64v^3$ , so the $u^2v$ and $uv^2$ coefficients are wrong.

Full solution

Row three is $1,3,3,1$ .
The term with $k$ copies of $-4v$ is $\binom3k(3u)^{3-k}(-4v)^k$ , giving $27u^3$ , $3\cdot9\cdot(-4)u^2v=-108u^2v$ , $3\cdot3\cdot16uv^2=144uv^2$ and $-64v^3$ .

The proposal has $-107$ and $143$ in place of $-108$ and $144$ .
At $u=v=1$ those two errors, $+1$ and $-1$ , cancel, so both sides give $-1$ although two coefficients are wrong.
A single substitution cannot establish an identity.

Answer

No; $(3u-4v)^3=27u^3-108u^2v+144uv^2-64v^3$ , so the $u^2v$ and $uv^2$ coefficients are wrong.

Key idea

A substitution check can miss coefficient errors that cancel at the chosen values.
Problem 8: Mark complete
Problem 9 Three conditions on a row

Advanced. This question goes beyond core Algebra II. It is not required by the course.

For a whole number $n$ , $(1+x)^n$ has a nonzero coefficient of $x^{17}$ , a zero coefficient of $x^{19}$ , and an odd coefficient of $x^2$ . Find $n$ and the coefficient of $x^2$ .
1. Hint 1
  For a whole number k, the coefficient of x to the k is nonzero exactly when k is at most n.
2. Hint 2
  The first two conditions leave two rows; compare their coefficients of x squared.
Answer

$n=18$ ; the coefficient of $x^2$ is $153$ .

Full solution

For whole numbers $k$ from $0$ through $n$ , the coefficient of $x^k$ in $(1+x)^n$ is $\binom nk$ , which is positive; for $k>n$ there is no $x^k$ term, so its coefficient is $0$ .
A nonzero $x^{17}$ coefficient needs $n\ge17$ , and a zero $x^{19}$ coefficient needs $n<19$ , so $n=17$ or $n=18$ .

Their $x^2$ coefficients are $\binom{17}2=136$ , which is even, and $\binom{18}2=153$ , which is odd.
So $n=18$ and the coefficient is $153$ .

Answer

$n=18$ ; the coefficient of $x^2$ is $153$ .

Key idea

Conditions on which coefficients are zero bound the exponent, and one more property can pick the row.
Problem 9: Mark complete
Problem 10 Two collected coefficients

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Let $P(x)=x^2(6+x)^3+(6+x)^4$ . A student claims that its $x^4$ coefficient is $19$ and its $x^6$ coefficient is $1$ . Evaluate both claims and identify the contributing term indices without expanding all of $P$ .
1. Hint 1
  The outside factor x squared changes the required power inside the first binomial.
2. Hint 2
  Find the target term index in each summand and check that it lies between zero and that summand's exponent.
Answer

The $x^4$ claim is correct, from $k=2$ in the first summand and $k=4$ in the second. The $x^6$ claim is false: that coefficient is $0$ , and neither summand has a contributing index.

Full solution

Write the terms as $\binom mk6^{m-k}x^k$ .
For $x^4$ , the first summand needs $x^2$ inside $(6+x)^3$ , so $k=2$ , contributing
$\binom32\cdot6=18.$
The second needs $k=4$ in $(6+x)^4$ , contributing $1$ .
Together the coefficient is $19$ .

For $x^6$ , the first summand would need $k=4$ , above its maximum of $3$ , and the second $k=6$ , above its maximum of $4$ .
Neither contributes, so the coefficient is $0$ , as the degree of $P$ , which is $5$ , also shows.

Answer

The $x^4$ claim is correct, from $k=2$ in the first summand and $k=4$ in the second. The $x^6$ claim is false: that coefficient is $0$ , and neither summand has a contributing index.

Key idea

A coefficient in a sum of binomial expressions collects every allowed contribution after accounting for outside factors.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra II. You can skip it. → Continue to Chapter 12: Conic Sections →

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