TIFR CSE 2021 | Part A | Question: 11

Question

Find the following sum.

$$\frac{1}{2^{2}-1}+\frac{1}{4^{2}-1}+\frac{1}{6^{2}-1}+\cdots+\frac{1}{40^{2}-1}$$

• A. $\frac{20}{41}$

• A. $\frac{10}{41}$

100. $\frac{10}{21}$

500. $\frac{20}{21}$

• A. $1$

Answer 1

Option $(A)$

$= \frac{1}{2^2-1} + \frac{1}{4^2-1} + ... + \frac{1}{40^2-1}$

$= \frac{1}{(2+1)(2-1)} + \frac{1}{(4+1)(4-1)} + ... +\frac{1}{(40+1)(40-1)}$

$= [\frac{1}{(2+1)(2-1)} + \frac{1}{(4+1)(4-1)} + ... +\frac{1}{(40+1)(40-1)}]\frac{2}{2}$

$= [\frac{2}{(2+1)(2-1)} + \frac{2}{(4+1)(4-1)} + ... +\frac{2}{(40+1)(40-1)}]\frac{1}{2}$

$= [\frac{(2+1)-(2-1)}{(2+1)(2-1)} + \frac{(4+1)-(4-1)}{(4+1)(4-1)} + ... +\frac{(40+1)-(40-1)}{(40+1)(40-1)}]\frac{1}{2}$

$= [\frac{1}{(2-1)} - \frac{1}{(2+1)} + \frac{1}{(4-1)} - \frac{1}{(4+1)} + ... +\frac{1}{(40-1)} - \frac{1}{(40+1)}]\frac{1}{2}$

$= [\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + ... +\frac{1}{39} - \frac{1}{41}]\frac{1}{2}$

$= [1 - \frac{1}{41}]\frac{1}{2}$

$= \frac{20}{41}$

Comments

How did you got an idea of multiplying and dividing the equation by 2?

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Usually, in these types of questions, if we play around a bit, we can cancel out the $2^{nd}$ term with the $3^{rd}$ term and $4^{th}$ term with the $5^{th}$ term and so on. And, at last, we are left with $1^{st}$ and the last term.

For that I had to somehow get $[\frac{1}{(2-1)}-\frac{1}{(2+1)}]$ from $\frac{1}{2^2-1}$ which I can get if I can convert $\frac{1}{2^2-1}$ into $\frac{(2+1)-(2-1)}{(2-1)(2+1)}$ (= $\frac{2}{(2-1)(2+1)}$).

So, I added $2$ in the numerator and denominator. Same thought with the other terms as well in the series. So, every term needs $2$ in the numerator to get spitted into two terms.