12 multiple-choice questions, progressively harder.
Find a closed form for ∑k=1n1(3k−2)(3k+1)\sum_{k=1}^{n} \frac{1}{(3k-2)(3k+1)}∑k=1n(3k−2)(3k+1)1.
Solution
Correct answer: D
The split is 13(13k−2−13k+1)\frac13\left(\frac{1}{3k-2} - \frac{1}{3k+1}\right)31(3k−21−3k+11), leaving 111 and −13n+1-\frac{1}{3n+1}−3n+11.
13(1−13n+1)=13⋅3n3n+1=n3n+1\frac13\left(1 - \frac{1}{3n+1}\right) = \frac13\cdot\frac{3n}{3n+1} = \frac{n}{3n+1}31(1−3n+11)=31⋅3n+13n=3n+1n
What is ∑k=1n(k+2−k)\sum_{k=1}^{n} \left(\sqrt{k+2} - \sqrt{k}\right)∑k=1n(k+2−k)?
Correct answer: B
This is a gap-2 telescope with bk=kb_k = \sqrt{k}bk=k, so two values survive at each end.
∑k=1n(k+2−k)=n+1+n+2−1−2\sum_{k=1}^{n}\left(\sqrt{k+2} - \sqrt{k}\right) = \sqrt{n+1} + \sqrt{n+2} - \sqrt1 - \sqrt2∑k=1n(k+2−k)=n+1+n+2−1−2
That is n+1+n+2−1−2\sqrt{n+1} + \sqrt{n+2} - 1 - \sqrt2n+1+n+2−1−2.
What is ∑k=17log2 (k+1k)\sum_{k=1}^{7} \log_{2}\!\left(\frac{k+1}{k}\right)∑k=17log2(kk+1)?
Correct answer: A
Each term is log2(k+1)−log2(k)\log_2(k+1) - \log_2(k)log2(k+1)−log2(k), so the sum telescopes to log2(n+1)\log_2(n+1)log2(n+1) with n=7n = 7n=7.
log2(8)=3\log_2(8) = 3log2(8)=3
Which value equals ∑k=1n[(k+1)3−k3]\sum_{k=1}^{n} \left[(k+1)^3 - k^3\right]∑k=1n[(k+1)3−k3]?
Correct answer: C
This is a telescope of consecutive cubes with bk=k3b_k = k^3bk=k3.
∑k=1n[(k+1)3−k3]=(n+1)3−13=(n+1)3−1\sum_{k=1}^{n}\left[(k+1)^3 - k^3\right] = (n+1)^3 - 1^3 = (n+1)^3 - 1∑k=1n[(k+1)3−k3]=(n+1)3−13=(n+1)3−1
What is ∑k=2101k(k+1)\sum_{k=2}^{10} \frac{1}{k(k+1)}∑k=210k(k+1)1 (the sum starts at k=2k = 2k=2)?
The first survivor is 12\frac1221 (from k=2k = 2k=2) and the last is −111-\frac{1}{11}−111 (from k=10k = 10k=10).
12−111=11−222=922\frac12 - \frac{1}{11} = \frac{11 - 2}{22} = \frac{9}{22}21−111=2211−2=229
What is the infinite sum ∑k=1∞1(4k−3)(4k+1)\sum_{k=1}^{\infty} \frac{1}{(4k-3)(4k+1)}∑k=1∞(4k−3)(4k+1)1?
The split is 14(14k−3−14k+1)\frac14\left(\frac{1}{4k-3} - \frac{1}{4k+1}\right)41(4k−31−4k+11), leaving only 111 at the front as the tail shrinks.
14(1−14n+1)→14\frac14\left(1 - \frac{1}{4n+1}\right) \to \frac1441(1−4n+11)→41
What is ∑k=124(k+1−k)\sum_{k=1}^{24} \left(\sqrt{k+1} - \sqrt{k}\right)∑k=124(k+1−k)?
The telescope collapses to n+1−1\sqrt{n+1} - 1n+1−1 with n=24n = 24n=24.
25−1=5−1=4\sqrt{25} - 1 = 5 - 1 = 425−1=5−1=4
In ∑k=1n1k(k+2)\sum_{k=1}^{n} \frac{1}{k(k+2)}∑k=1nk(k+2)1, why do the front survivors come out to 111 and 12\frac1221?
A gap of 222 means each negative term, −1k+2-\frac{1}{k+2}−k+21, cancels the matching +1k+2+\frac{1}{k+2}+k+21 two brackets later, so the first two positive terms, 111 and 12\frac1221, never cancel.
12(1+12−1n+1−1n+2)\frac12\left(1 + \frac12 - \frac{1}{n+1} - \frac{1}{n+2}\right)21(1+21−n+11−n+21)
Evaluate (11−12)+(12−13)+⋯+(149−150)\left(\frac11 - \frac12\right) + \left(\frac12 - \frac13\right) + \cdots + \left(\frac{1}{49} - \frac{1}{50}\right)(11−21)+(21−31)+⋯+(491−501).
The last bracket 149−150\frac{1}{49} - \frac{1}{50}491−501 means n=49n = 49n=49, so the survivors are 111 and −150-\frac{1}{50}−501.
1−150=49501 - \frac{1}{50} = \frac{49}{50}1−501=5049
What is ∑k=1991(2k−1)(2k+1)\sum_{k=1}^{99} \frac{1}{(2k-1)(2k+1)}∑k=199(2k−1)(2k+1)1?
The closed form is n2n+1\frac{n}{2n+1}2n+1n with n=99n = 99n=99.
992⋅99+1=99199\frac{99}{2\cdot 99 + 1} = \frac{99}{199}2⋅99+199=19999
For which nnn does ∑k=1n(k+1−k)=6\sum_{k=1}^{n} \left(\sqrt{k+1} - \sqrt{k}\right) = 6∑k=1n(k+1−k)=6?
The sum is n+1−1\sqrt{n+1} - 1n+1−1, so set it equal to 666.
n+1=7 ⇒ n+1=49 ⇒ n=48\sqrt{n+1} = 7 \;\Rightarrow\; n + 1 = 49 \;\Rightarrow\; n = 48n+1=7⇒n+1=49⇒n=48
What is the infinite sum ∑k=1∞(1k−1k+2)\sum_{k=1}^{\infty} \left(\frac1k - \frac{1}{k+2}\right)∑k=1∞(k1−k+21) (with no 12\frac1221 factor)?
With gap 222, the front survivors are 111 and 12\frac1221 while the tail shrinks away.
1+12−1n+1−1n+2→321 + \frac12 - \frac{1}{n+1} - \frac{1}{n+2} \to \frac321+21−n+11−n+21→23
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