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Question is to find the sum of:

$$(\frac{1}{2^2-1})+(\frac{1}{4^2-1})+(\frac{1}{6^2-1})+(\frac{1}{20^2-1})$$

I know that $a^2-b^2=(a+b)(a-b)$, and that with this I can find the LCM to be 1995, but this isn't helping me find the answer. I am looking for some kind of formula, because this question can come in an exam (having 60 qs for 60 mins), so they must not be giving a brute-force approach question.

Please help me in finding a fast approach.

Thanks.

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1 Answer 1

up vote 3 down vote accepted

It's also true that

$$\frac{1}{(a + b)(a - b)} = \frac{1}{2b}\left(\frac{1}{a - b} - \frac{1}{a + b}\right)$$

Thus you can write your sum as (choosing $b = 1$ in every term)

$$\frac 1 2 \left(\frac 1 1 - \frac 1 3 + \frac 1 3 - \frac 1 5 + \frac 1 5 - \frac 1 7 + \frac{1}{19} - \frac 1 {21}\right)$$

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So, I guuessed it right ! There's a technique for this problem. Thanks :) –  Gaurang Tandon May 5 at 16:56

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