12 multiple-choice questions, progressively harder.
Find a closed form for ∑k=1n1(2k−1)(2k+1)\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}∑k=1n(2k−1)(2k+1)1.
Solution
Correct answer: B
The split is 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), which telescopes so only 111 and −12n+1-\frac{1}{2n+1}−2n+11 survive.
12(1−12n+1)=12⋅2n2n+1=n2n+1\frac12\left(1 - \frac{1}{2n+1}\right) = \frac12\cdot\frac{2n}{2n+1} = \frac{n}{2n+1}21(1−2n+11)=21⋅2n+12n=2n+1n
What is the infinite sum 11⋅4+14⋅7+17⋅10+⋯\frac{1}{1\cdot 4} + \frac{1}{4\cdot 7} + \frac{1}{7\cdot 10} + \cdots1⋅41+4⋅71+7⋅101+⋯?
This is ∑k=1∞1(3k−2)(3k+1)\sum_{k=1}^{\infty}\frac{1}{(3k-2)(3k+1)}∑k=1∞(3k−2)(3k+1)1, whose finite total is n3n+1\frac{n}{3n+1}3n+1n.
n3n+1→13\frac{n}{3n+1} \to \frac133n+1n→31
The split 1k(k+3)=c(1k−1k+3)\frac{1}{k(k+3)} = c\left(\frac1k - \frac{1}{k+3}\right)k(k+3)1=c(k1−k+31) holds for what constant ccc?
Correct answer: A
Combining the bracket gives 1k−1k+3=(k+3)−kk(k+3)=3k(k+3)\frac1k - \frac{1}{k+3} = \frac{(k+3) - k}{k(k+3)} = \frac{3}{k(k+3)}k1−k+31=k(k+3)(k+3)−k=k(k+3)3, which is three times too big.
1k(k+3)=13(1k−1k+3)\frac{1}{k(k+3)} = \frac13\left(\frac1k - \frac{1}{k+3}\right)k(k+3)1=31(k1−k+31)
The triangular numbers are 1,3,6,10,15,…1, 3, 6, 10, 15, \ldots1,3,6,10,15,…, with Tk=k(k+1)2T_k = \frac{k(k+1)}{2}Tk=2k(k+1). What is 1T1+1T2+1T3+⋯\frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3} + \cdotsT11+T21+T31+⋯?
Correct answer: C
Each reciprocal is 1Tk=2k(k+1)=2(1k−1k+1)\frac{1}{T_k} = \frac{2}{k(k+1)} = 2\left(\frac1k - \frac{1}{k+1}\right)Tk1=k(k+1)2=2(k1−k+11), which telescopes.
2(1−1n+1)→22\left(1 - \frac{1}{n+1}\right) \to 22(1−n+11)→2
Find a closed form for ∑k=1n1(k+1)(k+2)\sum_{k=1}^{n} \frac{1}{(k+1)(k+2)}∑k=1n(k+1)(k+2)1.
Correct answer: D
The split is 1(k+1)(k+2)=1k+1−1k+2\frac{1}{(k+1)(k+2)} = \frac{1}{k+1} - \frac{1}{k+2}(k+1)(k+2)1=k+11−k+21, and the first survivor is 12\frac1221 (from k=1k = 1k=1).
∑k=1n(1k+1−1k+2)=12−1n+2\sum_{k=1}^{n}\left(\frac{1}{k+1} - \frac{1}{k+2}\right) = \frac12 - \frac{1}{n+2}∑k=1n(k+11−k+21)=21−n+21
What is 11+2+12+3+⋯+1120+121\frac{1}{\sqrt1 + \sqrt2} + \frac{1}{\sqrt2 + \sqrt3} + \cdots + \frac{1}{\sqrt{120} + \sqrt{121}}1+21+2+31+⋯+120+1211?
Each term rationalizes to k+1−k\sqrt{k+1} - \sqrt{k}k+1−k, and the last denominator uses k=120k = 120k=120, so the sum telescopes to 121−1\sqrt{121} - 1121−1.
121−1=11−1=10\sqrt{121} - 1 = 11 - 1 = 10121−1=11−1=10
What is the infinite sum 12⋅3+13⋅4+14⋅5+⋯\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \frac{1}{4\cdot 5} + \cdots2⋅31+3⋅41+4⋅51+⋯?
This is ∑k=1∞1(k+1)(k+2)\sum_{k=1}^{\infty}\frac{1}{(k+1)(k+2)}∑k=1∞(k+1)(k+2)1, whose finite total is 12−1n+2\frac12 - \frac{1}{n+2}21−n+21.
12−1n+2→12\frac12 - \frac{1}{n+2} \to \frac1221−n+21→21
What is 13⋅5+15⋅7+17⋅9+⋯+199⋅101\frac{1}{3\cdot 5} + \frac{1}{5\cdot 7} + \frac{1}{7\cdot 9} + \cdots + \frac{1}{99\cdot 101}3⋅51+5⋅71+7⋅91+⋯+99⋅1011?
Each term is 12(12m+1−12m+3)\frac12\left(\frac{1}{2m+1} - \frac{1}{2m+3}\right)21(2m+11−2m+31); the first survivor is 13\frac1331 and the last is −1101-\frac{1}{101}−1011.
12(13−1101)=12⋅98303=49303\frac12\left(\frac13 - \frac{1}{101}\right) = \frac12\cdot\frac{98}{303} = \frac{49}{303}21(31−1011)=21⋅30398=30349
Which infinite telescoping sum diverges (has no finite value)?
That sum rationalizes to k+1−k\sqrt{k+1} - \sqrt{k}k+1−k, whose total n+1−1\sqrt{n+1} - 1n+1−1 grows without bound.
n+1−1→∞\sqrt{n+1} - 1 \to \inftyn+1−1→∞
The others settle at 34\frac3443, 12\frac1221, and 222.
What is the infinite sum 12⋅4+14⋅6+16⋅8+⋯\frac{1}{2\cdot 4} + \frac{1}{4\cdot 6} + \frac{1}{6\cdot 8} + \cdots2⋅41+4⋅61+6⋅81+⋯?
The general term is 1(2k)(2k+2)=14k(k+1)=14(1k−1k+1)\frac{1}{(2k)(2k+2)} = \frac{1}{4k(k+1)} = \frac14\left(\frac1k - \frac{1}{k+1}\right)(2k)(2k+2)1=4k(k+1)1=41(k1−k+11), which telescopes.
14(1−1n+1)→14\frac14\left(1 - \frac{1}{n+1}\right) \to \frac1441(1−n+11)→41
Evaluate ∑k=1nln (1+1k)\sum_{k=1}^{n} \ln\!\left(1 + \frac1k\right)∑k=1nln(1+k1).
Rewrite 1+1k=k+1k1 + \frac1k = \frac{k+1}{k}1+k1=kk+1, so each term is ln(k+1)−ln(k)\ln(k+1) - \ln(k)ln(k+1)−ln(k), a telescope.
∑k=1n[ln(k+1)−ln(k)]=ln(n+1)−ln(1)=ln(n+1)\sum_{k=1}^{n}\big[\ln(k+1) - \ln(k)\big] = \ln(n+1) - \ln(1) = \ln(n+1)∑k=1n[ln(k+1)−ln(k)]=ln(n+1)−ln(1)=ln(n+1)
Which infinite sum does NOT telescope to a finite value?
The square-root sum telescopes to n+1−1\sqrt{n+1} - 1n+1−1, whose surviving tail grows without bound.
The first three settle at 111, 12\frac1221, and 222.
Reset this practice set?
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