12 multiple-choice questions, progressively harder.
Which is the correct split of 1k(k+1)\frac{1}{k(k+1)}k(k+1)1 into a difference of two fractions?
Solution
Correct answer: C
Check a split by combining it over a common denominator and seeing whether it rebuilds the original.
1k−1k+1=(k+1)−kk(k+1)=1k(k+1)\frac{1}{k} - \frac{1}{k+1} = \frac{(k+1) - k}{k(k+1)} = \frac{1}{k(k+1)}k1−k+11=k(k+1)(k+1)−k=k(k+1)1
The two sides agree, so this is the correct split.
Combine 1k−1k+1\frac{1}{k} - \frac{1}{k+1}k1−k+11 into a single fraction.
Correct answer: D
Use the common denominator k(k+1)k(k+1)k(k+1) and subtract the numerators.
Evaluate (11−12)+(12−13)+(13−14)\left(\frac11 - \frac12\right) + \left(\frac12 - \frac13\right) + \left(\frac13 - \frac14\right)(11−21)+(21−31)+(31−41).
Correct answer: B
The −12-\frac12−21 cancels the +12+\frac12+21, and the −13-\frac13−31 cancels the +13+\frac13+31. Only the first and last pieces survive.
1−14=341 - \frac14 = \frac341−41=43
Expand (k+1)2−k2(k+1)^2 - k^2(k+1)2−k2.
Expand the square and subtract k2k^2k2.
(k+1)2−k2=k2+2k+1−k2=2k+1(k+1)^2 - k^2 = k^2 + 2k + 1 - k^2 = 2k + 1(k+1)2−k2=k2+2k+1−k2=2k+1
In (11−12)+(12−13)+⋯+(1n−1n+1)\left(\frac11 - \frac12\right) + \left(\frac12 - \frac13\right) + \cdots + \left(\frac1n - \frac{1}{n+1}\right)(11−21)+(21−31)+⋯+(n1−n+11), which two pieces survive?
The front 111 is never subtracted and the final −1n+1-\frac{1}{n+1}−n+11 is never added back; everything between them cancels.
∑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 the value of the infinite sum 11⋅2+12⋅3+13⋅4+⋯\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \cdots1⋅21+2⋅31+3⋅41+⋯?
The finite total is 1−1n+11 - \frac{1}{n+1}1−n+11, and the tail 1n+1\frac{1}{n+1}n+11 shrinks toward zero as nnn grows.
1−1n+1→11 - \frac{1}{n+1} \to 11−n+11→1
So the running total settles at 111.
What is ∑k=1n(k+1−k)\sum_{k=1}^{n} \left(\sqrt{k+1} - \sqrt{k}\right)∑k=1n(k+1−k)?
This is a telescope with bk=kb_k = \sqrt{k}bk=k in the mirror direction, so the last value survives positive and the first negative.
∑k=1n(k+1−k)=n+1−1=n+1−1\sum_{k=1}^{n} \left(\sqrt{k+1} - \sqrt{k}\right) = \sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1∑k=1n(k+1−k)=n+1−1=n+1−1
To confirm that 1k(k+1)=1k−1k+1\frac{1}{k(k+1)} = \frac1k - \frac{1}{k+1}k(k+1)1=k1−k+11 is a valid split, what should you do?
Correct answer: A
Recombining the right side should rebuild the left side exactly.
1k−1k+1=(k+1)−kk(k+1)=1k(k+1)\frac1k - \frac{1}{k+1} = \frac{(k+1) - k}{k(k+1)} = \frac{1}{k(k+1)}k1−k+11=k(k+1)(k+1)−k=k(k+1)1
Because it matches, the split is safe to use.
What is ∑k=121k(k+1)\sum_{k=1}^{2} \frac{1}{k(k+1)}∑k=12k(k+1)1?
Add the two terms 11⋅2\frac{1}{1\cdot 2}1⋅21 and 12⋅3\frac{1}{2\cdot 3}2⋅31, or use 1−1n+11 - \frac{1}{n+1}1−n+11 with n=2n = 2n=2.
1−13=231 - \frac{1}{3} = \frac231−31=32
A telescope from k=1k = 1k=1 to k=4k = 4k=4 leaves the survivors 111 and −15-\frac15−51. What is the sum?
Add the two surviving pieces.
1−15=451 - \frac15 = \frac451−51=54
Which statement best describes a telescoping sum?
A telescoping sum is built so that each term is a difference bk−bk+1b_k - b_{k+1}bk−bk+1.
∑k=1n(bk−bk+1)=b1−bn+1\sum_{k=1}^{n}(b_k - b_{k+1}) = b_1 - b_{n+1}∑k=1n(bk−bk+1)=b1−bn+1
The interior cancels in pairs and only the two ends survive.
A mirror telescope has ak=bk+1−bka_k = b_{k+1} - b_kak=bk+1−bk. What is ∑k=1nak\sum_{k=1}^{n} a_k∑k=1nak?
The same pairing cancels the interior, but now the last value survives positive and the first negative.
∑k=1n(bk+1−bk)=bn+1−b1\sum_{k=1}^{n}(b_{k+1} - b_k) = b_{n+1} - b_1∑k=1n(bk+1−bk)=bn+1−b1
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.