12 multiple-choice questions, progressively harder.
What is the infinite sum 11⋅3+13⋅5+15⋅7+⋯\frac{1}{1\cdot 3} + \frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \cdots1⋅31+3⋅51+5⋅71+⋯?
Solution
Correct answer: D
The finite total is n2n+1\frac{n}{2n+1}2n+1n, and as nnn grows this approaches 12\frac1221.
n2n+1→12\frac{n}{2n+1} \to \frac122n+1n→21
Evaluate ∑k=1nlog (k+1k)\sum_{k=1}^{n} \log\!\left(\frac{k+1}{k}\right)∑k=1nlog(kk+1).
Correct answer: A
Split the log of a quotient: log (k+1k)=log(k+1)−log(k)\log\!\left(\frac{k+1}{k}\right) = \log(k+1) - \log(k)log(kk+1)=log(k+1)−log(k), a telescope with bk=log(k)b_k = \log(k)bk=log(k).
∑k=1n[log(k+1)−log(k)]=log(n+1)−log(1)=log(n+1)\sum_{k=1}^{n}\big[\log(k+1) - \log(k)\big] = \log(n+1) - \log(1) = \log(n+1)∑k=1n[log(k+1)−log(k)]=log(n+1)−log(1)=log(n+1)
What is ∑k=19log10 (k+1k)\sum_{k=1}^{9} \log_{10}\!\left(\frac{k+1}{k}\right)∑k=19log10(kk+1)?
Correct answer: B
The sum telescopes to log10(n+1)\log_{10}(n+1)log10(n+1) with n=9n = 9n=9.
log10(10)=1\log_{10}(10) = 1log10(10)=1
For which nnn does ∑k=1n1k(k+1)=910\sum_{k=1}^{n} \frac{1}{k(k+1)} = \frac{9}{10}∑k=1nk(k+1)1=109?
Set the collapsed total 1−1n+11 - \frac{1}{n+1}1−n+11 equal to 910\frac{9}{10}109.
1−1n+1=910 ⇒ 1n+1=110 ⇒ n=91 - \frac{1}{n+1} = \frac{9}{10} \;\Rightarrow\; \frac{1}{n+1} = \frac{1}{10} \;\Rightarrow\; n = 91−n+11=109⇒n+11=101⇒n=9
What is ∑k=1801k+k+1\sum_{k=1}^{80} \frac{1}{\sqrt{k} + \sqrt{k+1}}∑k=180k+k+11?
Each term rationalizes to k+1−k\sqrt{k+1} - \sqrt{k}k+1−k, so the sum telescopes to n+1−1\sqrt{n+1} - 1n+1−1 with n=80n = 80n=80.
81−1=9−1=8\sqrt{81} - 1 = 9 - 1 = 881−1=9−1=8
What is ∑k=1n1k+k+1\sum_{k=1}^{n} \frac{1}{\sqrt{k} + \sqrt{k+1}}∑k=1nk+k+11?
Rationalizing each term gives k+1−k\sqrt{k+1} - \sqrt{k}k+1−k, which telescopes.
∑k=1n(k+1−k)=n+1−1\sum_{k=1}^{n}\left(\sqrt{k+1} - \sqrt{k}\right) = \sqrt{n+1} - 1∑k=1n(k+1−k)=n+1−1
For which nnn is ∑k=1n(2k+1)=224\sum_{k=1}^{n} (2k+1) = 224∑k=1n(2k+1)=224?
Correct answer: C
The sum telescopes to (n+1)2−1(n+1)^2 - 1(n+1)2−1, so set it equal to 224224224.
(n+1)2−1=224 ⇒ (n+1)2=225 ⇒ n+1=15 ⇒ n=14(n+1)^2 - 1 = 224 \;\Rightarrow\; (n+1)^2 = 225 \;\Rightarrow\; n + 1 = 15 \;\Rightarrow\; n = 14(n+1)2−1=224⇒(n+1)2=225⇒n+1=15⇒n=14
What is ∑k=1501(2k−1)(2k+1)\sum_{k=1}^{50} \frac{1}{(2k-1)(2k+1)}∑k=150(2k−1)(2k+1)1?
The closed form is n2n+1\frac{n}{2n+1}2n+1n with n=50n = 50n=50.
502⋅50+1=50101\frac{50}{2\cdot 50 + 1} = \frac{50}{101}2⋅50+150=10150
What is ∑k=5991k(k+1)\sum_{k=5}^{99} \frac{1}{k(k+1)}∑k=599k(k+1)1?
The first survivor is 15\frac1551 (from k=5k = 5k=5) and the last is −1100-\frac{1}{100}−1001 (from k=99k = 99k=99).
15−1100=20−1100=19100\frac15 - \frac{1}{100} = \frac{20 - 1}{100} = \frac{19}{100}51−1001=10020−1=10019
Which is the correct split of 1(2k−1)(2k+1)\frac{1}{(2k-1)(2k+1)}(2k−1)(2k+1)1?
Combining 12k−1−12k+1\frac{1}{2k-1} - \frac{1}{2k+1}2k−11−2k+11 gives 2(2k−1)(2k+1)\frac{2}{(2k-1)(2k+1)}(2k−1)(2k+1)2, so the split needs a factor of 12\frac1221.
1(2k−1)(2k+1)=12(12k−1−12k+1)\frac{1}{(2k-1)(2k+1)} = \frac12\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)(2k−1)(2k+1)1=21(2k−11−2k+11)
What is the infinite sum ∑k=1∞1k(k+3)\sum_{k=1}^{\infty} \frac{1}{k(k+3)}∑k=1∞k(k+3)1?
With gap 333, the split 13(1k−1k+3)\frac13\left(\frac1k - \frac{1}{k+3}\right)31(k1−k+31) leaves three front survivors 11,12,13\frac11, \frac12, \frac1311,21,31; the tail shrinks away.
13(1+12+13)=13⋅116=1118\frac13\left(1 + \frac12 + \frac13\right) = \frac13\cdot\frac{11}{6} = \frac{11}{18}31(1+21+31)=31⋅611=1811
A student claims ∑k=1n1k(k+2)=1−1n+2\sum_{k=1}^{n} \frac{1}{k(k+2)} = 1 - \frac{1}{n+2}∑k=1nk(k+2)1=1−n+21, keeping one survivor at each end. What is the correct closed form?
A gap-2 telescope leaves two survivors at each end, and the split carries a factor of 12\frac1221.
12(1+12−1n+1−1n+2)=34−12(1n+1+1n+2)\frac12\left(1 + \frac12 - \frac{1}{n+1} - \frac{1}{n+2}\right) = \frac34 - \frac12\left(\frac{1}{n+1} + \frac{1}{n+2}\right)21(1+21−n+11−n+21)=43−21(n+11+n+21)
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