Difference between revisions of "Telescoping series"
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==Problems== | ==Problems== | ||
===Introductory=== | ===Introductory=== | ||
+ | *When simplified the product <math>\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)</math> becomes: | ||
+ | <math>\textbf{(A)}\ \frac1n \qquad\textbf{(B)}\ \frac2n\qquad\textbf{(C)}\ \frac{2(n-1)}{n}\qquad\textbf{(D)}\ \frac{2}{n(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)} </math> ([[1959 AHSME Problems/Problem 37|Source]]) | ||
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*The sum <math>\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}</math> can be expressed as <math>a-\frac{1}{b!}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>? ([[2022 AMC 10B Problems/Problem 9|Source]]) | *The sum <math>\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}</math> can be expressed as <math>a-\frac{1}{b!}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>? ([[2022 AMC 10B Problems/Problem 9|Source]]) | ||
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([[2021 AMC 12A Problems/Problem 9|Source]]) | ([[2021 AMC 12A Problems/Problem 9|Source]]) | ||
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===Intermediate=== | ===Intermediate=== |
Latest revision as of 15:37, 21 July 2024
In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. This is often done by using a form of for some expression .
Contents
[hide]Example 1
Derive the formula for the sum of the first counting numbers.
Solution 1
We wish to write for some expression . This expression is as .
We then telescope the expression:
.
(Notice how the sum telescopes— contains a positive and a negative of every value of from to , so those terms cancel. We are then left with , the only terms which did not cancel.)
Example 2
Find a general formula for , where .
Solution 2
We wish to write for some expression . This can be easily achieved with as by simple computation.
We then telescope the expression:
.
Problems
Introductory
- When simplified the product becomes:
(Source)
- The sum can be expressed as , where and are positive integers. What is ? (Source)
- Which of the following is equivalent to (Hint: difference of squares!)
(Source)
Intermediate
- Let denote the value of the sum can be expressed as , where and are positive integers and is not divisible by the square of any prime. Determine . (Source)
Olympiad
- Find the value of , where is the Riemann zeta function