12 multiple-choice questions, progressively harder.
Evaluate ∑k=24(1k−1k+1)\displaystyle\sum_{k=2}^{4} \left( \frac{1}{k} - \frac{1}{k+1} \right)k=2∑4(k1−k+11).
Solution
Correct answer: D
Write out the three terms and watch what happens in the middle.
(12−13)+(13−14)+(14−15)\left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right)(21−31)+(31−41)+(41−51)
The −13-\frac{1}{3}−31 cancels the +13+\frac{1}{3}+31, and the −14-\frac{1}{4}−41 cancels the +14+\frac{1}{4}+41, so only the very first and very last pieces survive.
12−15=510−210=310\frac{1}{2} - \frac{1}{5} = \frac{5}{10} - \frac{2}{10} = \frac{3}{10}21−51=105−102=103
Evaluate ∑k=25(−1)kk2\displaystyle\sum_{k=2}^{5} (-1)^k k^2k=2∑5(−1)kk2.
Correct answer: B
The factor (−1)k(-1)^k(−1)k is +1+1+1 at even indices and −1-1−1 at odd ones, and the sum starts at the even index k=2k = 2k=2.
(+4)+(−9)+(+16)+(−25)(+4) + (-9) + (+16) + (-25)(+4)+(−9)+(+16)+(−25)
Add left to right.
4−9+16−25=−144 - 9 + 16 - 25 = -144−9+16−25=−14
Ignoring the sign factor gives 545454, which is the sum of the squares alone.
A sequence has a1=3a_1 = 3a1=3, a2=4a_2 = 4a2=4, and an=an−1−an−2a_n = a_{n-1} - a_{n-2}an=an−1−an−2 for n≥3n \ge 3n≥3. What is a20a_{20}a20?
Correct answer: A
Generate terms until the pattern gives itself away.
3, 4, 1, −3, −4, −1, 3, 4, …3,\ 4,\ 1,\ -3,\ -4,\ -1,\ 3,\ 4,\ \ldots3, 4, 1, −3, −4, −1, 3, 4, …
By a7a_7a7 and a8a_8a8 the pair of seeds has returned, so the sequence repeats with period 666, meaning an+6=ana_{n+6} = a_nan+6=an for every nnn. Now reduce the index.
20=6(3)+2 ⟹ a20=a2=420 = 6(3) + 2 \;\Longrightarrow\; a_{20} = a_2 = 420=6(3)+2⟹a20=a2=4
Evaluate ∑j=77j2\displaystyle\sum_{j=7}^{7} j^2j=7∑7j2.
The lower and upper limits are equal, so count the terms the usual way.
7−7+1=1 term7 - 7 + 1 = 1 \text{ term}7−7+1=1 term
The index takes the single value j=7j = 7j=7, and the sum is just that one term.
∑j=77j2=72=49\sum_{j=7}^{7} j^2 = 7^2 = 49∑j=77j2=72=49
The answer 111 is the number of terms, not the value of the sum.
A sequence is given by a1=2a_1 = 2a1=2 and an=an−1+7a_n = a_{n-1} + 7an=an−1+7 for n≥2n \ge 2n≥2. Using only this rule, how many additions must you perform to reach a1000a_{1000}a1000?
Each use of the rule performs one addition and advances the index by one, and the rule is used at n=2,3,4,…,1000n = 2, 3, 4, \ldots, 1000n=2,3,4,…,1000.
1000−2+1=9991000 - 2 + 1 = 9991000−2+1=999
So reaching a1000a_{1000}a1000 costs 999999999 additions. This is precisely the weakness of a recursive rule, and precisely the reason an explicit rule is worth finding: one would hand you a1000a_{1000}a1000 in a single substitution.
A sequence has a1=1a_1 = 1a1=1, a2=2a_2 = 2a2=2, and an=an−1⋅an−2+1a_n = a_{n-1} \cdot a_{n-2} + 1an=an−1⋅an−2+1 for n≥3n \ge 3n≥3. What is a5a_5a5?
The rule multiplies the two previous terms and then adds 111, so both seeds are needed and every term must be built in order.
a3=a2⋅a1+1=2(1)+1=3,a4=a3⋅a2+1=3(2)+1=7a_3 = a_2 \cdot a_1 + 1 = 2(1) + 1 = 3, \qquad a_4 = a_3 \cdot a_2 + 1 = 3(2) + 1 = 7a3=a2⋅a1+1=2(1)+1=3,a4=a3⋅a2+1=3(2)+1=7
Now use a4a_4a4 and a3a_3a3.
a5=a4⋅a3+1=7(3)+1=22a_5 = a_4 \cdot a_3 + 1 = 7(3) + 1 = 22a5=a4⋅a3+1=7(3)+1=22
How many terms are in the sum ∑j=620j2\displaystyle\sum_{j=6}^{20} j^2j=6∑20j2?
Correct answer: C
Count the indices, not the steps between them, and remember that both limits are included.
20−6+1=1520 - 6 + 1 = 1520−6+1=15
The index takes the values 6,7,8,…,206, 7, 8, \ldots, 206,7,8,…,20, which is 151515 values. The summand being j2j^2j2 rather than jjj makes no difference to the count.
Which sum is equal to ∑k=14(k+1)3\displaystyle\sum_{k=1}^{4} (k+1)^3k=1∑4(k+1)3?
As kkk runs from 111 to 444, the quantity k+1k + 1k+1 runs from 222 to 555, so the numbers being cubed are 2,3,4,52, 3, 4, 52,3,4,5.
∑k=14(k+1)3=8+27+64+125=224\sum_{k=1}^{4} (k+1)^3 = 8 + 27 + 64 + 125 = 224∑k=14(k+1)3=8+27+64+125=224
Substituting j=k+1j = k + 1j=k+1 raises both limits by one and clears the +1+1+1 from the summand, giving ∑j=25j3\sum_{j=2}^{5} j^3∑j=25j3. Shifting the limits and keeping the +1+1+1 double-counts the shift: the summand (j+1)3(j+1)^3(j+1)3 run from j=2j = 2j=2 to j=5j = 5j=5 cubes 3,4,5,63, 4, 5, 63,4,5,6 instead, giving 432432432.
Which sum has the value 353535?
Expand each candidate rather than guessing, since the limits matter as much as the summand.
∑k=15(2k+1)=3+5+7+9+11=35\sum_{k=1}^{5} (2k+1) = 3 + 5 + 7 + 9 + 11 = 35∑k=15(2k+1)=3+5+7+9+11=35
The others fall short: ∑k=152k=30\sum_{k=1}^{5} 2k = 30∑k=152k=30 drops the +1+1+1 from every term, ∑k=14(2k+1)=24\sum_{k=1}^{4} (2k+1) = 24∑k=14(2k+1)=24 stops one term early, and ∑k=25(2k+1)=32\sum_{k=2}^{5} (2k+1) = 32∑k=25(2k+1)=32 starts one term late.
The sums ∑i=14i2\displaystyle\sum_{i=1}^{4} i^2i=1∑4i2 and ∑k=14k2\displaystyle\sum_{k=1}^{4} k^2k=1∑4k2 have the same value. What is the reason?
The index of summation is a dummy: it exists only to walk through the values between the limits, and it never survives into the answer.
∑i=14i2=1+4+9+16=30=∑k=14k2\sum_{i=1}^{4} i^2 = 1 + 4 + 9 + 16 = 30 = \sum_{k=1}^{4} k^2∑i=14i2=1+4+9+16=30=∑k=14k2
Both notations issue the same instruction, so both produce the same list of terms. The only caution is that the new name must not already be doing another job in the expression. The claim that only the upper limit matters is false, since changing the lower limit or the summand certainly changes the value.
A jar contains 505050 dollars. At the end of every month you add 121212 dollars. Let ana_nan be the amount in the jar at the end of month nnn. Which recursive definition describes the sequence?
Find the seed first. At the end of the first month the jar holds the original amount plus one deposit.
a1=50+12=62a_1 = 50 + 12 = 62a1=50+12=62
Each later month adds one more deposit to whatever was there, which is the step rule an=an−1+12a_n = a_{n-1} + 12an=an−1+12. The option with no seed describes infinitely many sequences, one for each starting amount, so it does not define this one.
Evaluate ∑k=14kk+1\displaystyle\sum_{k=1}^{4} \frac{k}{k+1}k=1∑4k+1k.
Write out the four terms, then add them over a common denominator. Fractions are not added by adding numerators and denominators separately.
12+23+34+45\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5}21+32+43+54
The least common denominator of 2,3,4,52, 3, 4, 52,3,4,5 is 606060.
3060+4060+4560+4860=16360\frac{30}{60} + \frac{40}{60} + \frac{45}{60} + \frac{48}{60} = \frac{163}{60}6030+6040+6045+6048=60163
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