Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
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== Solution 2 == | == Solution 2 == | ||
+ | |||
+ | Simplifying the expression yields | ||
+ | <cmath>\begin{align*} | ||
+ | S &= \sum_{n=1}^{9800}\frac{1}{\sqrt{n+\sqrt{n^1-1}}} \ | ||
+ | &= \sum_{n=1}^{9800}\frac{\sqrt{n+\sqrt{n^1-1}}}{n+\sqrt{n^1-1}} \ | ||
+ | &= \sum_{n=1}^{9800}(\frac{\sqrt{n+\sqrt{n^1-1}}}{n+\sqrt{n^1-1}})\cdot(\frac{n-\sqrt{n^2-1}}{n-\sqrt{n^2-1}}) \ | ||
+ | &= \sum_{n=1}^{9800}\frac{(\sqrt{n+\sqrt{n^1-1})\cdot(\sqrt{n+\sqrt{n^1-1})^2}{\sqrt{n+\sqrt{n^1-1}}} \ | ||
+ | \end{align*}</cmath> | ||
==See Also== | ==See Also== |
Revision as of 10:11, 28 November 2019
Contents
[hide]Problem
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 .
Solution
Notice that . Thus, we have
This is a telescoping series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with , and .
Solution 2
Simplifying the expression yields
\begin{align*} S &= \sum_{n=1}^{9800}\frac{1}{\sqrt{n+\sqrt{n^1-1}}} \\ &= \sum_{n=1}^{9800}\frac{\sqrt{n+\sqrt{n^1-1}}}{n+\sqrt{n^1-1}} \\ &= \sum_{n=1}^{9800}(\frac{\sqrt{n+\sqrt{n^1-1}}}{n+\sqrt{n^1-1}})\cdot(\frac{n-\sqrt{n^2-1}}{n-\sqrt{n^2-1}}) \\ &= \sum_{n=1}^{9800}\frac{(\sqrt{n+\sqrt{n^1-1})\cdot(\sqrt{n+\sqrt{n^1-1})^2}{\sqrt{n+\sqrt{n^1-1}}} \\ \end{align*} (Error compiling LaTeX. Unknown error_msg)
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |