12 multiple-choice questions, progressively harder.
What is ∑k=1201k(k+1)\sum_{k=1}^{20} \frac{1}{k(k+1)}∑k=120k(k+1)1?
Solution
Correct answer: B
The telescope collapses to 1−1n+11 - \frac{1}{n+1}1−n+11, so use n=20n = 20n=20.
1−121=20211 - \frac{1}{21} = \frac{20}{21}1−211=2120
What is the value of the infinite sum 11⋅3+12⋅4+13⋅5+⋯\frac{1}{1\cdot 3} + \frac{1}{2\cdot 4} + \frac{1}{3\cdot 5} + \cdots1⋅31+2⋅41+3⋅51+⋯?
Correct answer: C
Each term is 1k(k+2)=12(1k−1k+2)\frac{1}{k(k+2)} = \frac12\left(\frac1k - \frac{1}{k+2}\right)k(k+2)1=21(k1−k+21), and the finite total is 34−12(1n+1+1n+2)\frac34 - \frac12\left(\frac{1}{n+1} + \frac{1}{n+2}\right)43−21(n+11+n+21). Both leftover terms shrink to zero.
34−12(1n+1+1n+2)→34\frac34 - \frac12\left(\frac{1}{n+1} + \frac{1}{n+2}\right) \to \frac3443−21(n+11+n+21)→43
Which is the correct split of 1k(k+2)\frac{1}{k(k+2)}k(k+2)1?
Combining 1k−1k+2\frac1k - \frac{1}{k+2}k1−k+21 gives 2k(k+2)\frac{2}{k(k+2)}k(k+2)2, which is twice too big, so the split needs a factor of 12\frac1221.
1k−1k+2=2k(k+2) ⇒ 1k(k+2)=12(1k−1k+2)\frac1k - \frac{1}{k+2} = \frac{2}{k(k+2)} \;\Rightarrow\; \frac{1}{k(k+2)} = \frac12\left(\frac1k - \frac{1}{k+2}\right)k1−k+21=k(k+2)2⇒k(k+2)1=21(k1−k+21)
What is ∑k=148(k+1−k)\sum_{k=1}^{48} \left(\sqrt{k+1} - \sqrt{k}\right)∑k=148(k+1−k)?
Correct answer: A
The telescope collapses to n+1−1\sqrt{n+1} - 1n+1−1 with n=48n = 48n=48.
49−1=7−1=6\sqrt{49} - 1 = 7 - 1 = 649−1=7−1=6
What is ∑k=1n(2k+1)\sum_{k=1}^{n} (2k+1)∑k=1n(2k+1), using the split 2k+1=(k+1)2−k22k+1 = (k+1)^2 - k^22k+1=(k+1)2−k2?
With bk=k2b_k = k^2bk=k2 the sum telescopes to the last square minus the first.
∑k=1n(2k+1)=(n+1)2−12=(n+1)2−1\sum_{k=1}^{n}(2k+1) = (n+1)^2 - 1^2 = (n+1)^2 - 1∑k=1n(2k+1)=(n+1)2−12=(n+1)2−1
In the gap-2 sum ∑k=1n1k(k+2)\sum_{k=1}^{n} \frac{1}{k(k+2)}∑k=1nk(k+2)1, which pieces survive at the front end?
Correct answer: D
With a gap of 222, each negative cancels the positive two brackets later, so the first two positives 11\frac1111 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)
Using ∑k=1n1k(k+2)=34−12(1n+1+1n+2)\sum_{k=1}^{n} \frac{1}{k(k+2)} = \frac34 - \frac12\left(\frac{1}{n+1} + \frac{1}{n+2}\right)∑k=1nk(k+2)1=43−21(n+11+n+21), what is the sum for n=5n = 5n=5?
Substitute n=5n = 5n=5, so the leftover terms are 16\frac1661 and 17\frac1771.
34−12(16+17)=34−12⋅1342=34−1384=2542\frac34 - \frac12\left(\frac16 + \frac17\right) = \frac34 - \frac12\cdot\frac{13}{42} = \frac34 - \frac{13}{84} = \frac{25}{42}43−21(61+71)=43−21⋅4213=43−8413=4225
The finite sum is ∑k=1n1k(k+1)=1−1n+1\sum_{k=1}^{n} \frac{1}{k(k+1)} = 1 - \frac{1}{n+1}∑k=1nk(k+1)1=1−n+11. What is its value for n=9n = 9n=9?
Set n=9n = 9n=9.
1−110=9101 - \frac{1}{10} = \frac{9}{10}1−101=109
The value 111 is only the infinite sum, not the total at n=9n = 9n=9.
For which nnn does ∑k=1n(k+1−k)=4\sum_{k=1}^{n}\left(\sqrt{k+1} - \sqrt{k}\right) = 4∑k=1n(k+1−k)=4?
The sum is n+1−1\sqrt{n+1} - 1n+1−1, so set it equal to 444.
n+1−1=4 ⇒ n+1=5 ⇒ n+1=25 ⇒ n=24\sqrt{n+1} - 1 = 4 \;\Rightarrow\; \sqrt{n+1} = 5 \;\Rightarrow\; n + 1 = 25 \;\Rightarrow\; n = 24n+1−1=4⇒n+1=5⇒n+1=25⇒n=24
Combine 12(1k−1k+2)\frac12\left(\frac1k - \frac{1}{k+2}\right)21(k1−k+21) into a single fraction.
Inside the bracket, 1k−1k+2=2k(k+2)\frac1k - \frac{1}{k+2} = \frac{2}{k(k+2)}k1−k+21=k(k+2)2; the outer 12\frac1221 halves it.
12⋅2k(k+2)=1k(k+2)\frac12\cdot\frac{2}{k(k+2)} = \frac{1}{k(k+2)}21⋅k(k+2)2=k(k+2)1
What is ∑k=1n2k(k+1)\sum_{k=1}^{n} \frac{2}{k(k+1)}∑k=1nk(k+1)2?
Factor out the 222: 2k(k+1)=2(1k−1k+1)\frac{2}{k(k+1)} = 2\left(\frac1k - \frac{1}{k+1}\right)k(k+1)2=2(k1−k+11), and the bracket telescopes to 1−1n+11 - \frac{1}{n+1}1−n+11.
2(1−1n+1)=2nn+12\left(1 - \frac{1}{n+1}\right) = \frac{2n}{n+1}2(1−n+11)=n+12n
What is the value of the infinite sum ∑k=1∞2k(k+1)\sum_{k=1}^{\infty} \frac{2}{k(k+1)}∑k=1∞k(k+1)2?
The finite total is 2nn+1=2(1−1n+1)\frac{2n}{n+1} = 2\left(1 - \frac{1}{n+1}\right)n+12n=2(1−n+11), and the tail shrinks to zero.
2(1−1n+1)→22\left(1 - \frac{1}{n+1}\right) \to 22(1−n+11)→2
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