Question

$\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\cdots \cdots +\frac{2}{(2n-1)(2n+1)}$ is equal to

• A. $\frac{2n}{2n-1}$
• B. $\frac{2n}{2n+1}$
• C. $\frac{2n}{2n-3}$
• D. $\frac{2n}{2n+3}$

Answer 1

The correct option is B $\frac{2n}{2n+1}$

$S_n=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+......+\frac{2}{(2n-1)(2n+1)}$

General $n^{th}$ term can be written as

$T_n=\frac{2}{(2n-1)(2n+1)}$

$S_n=\sum _{n=1}^n\frac{2}{(2n-1)(2n+1)}$

Now, $S_n$ can also be written as

$S_n=\sum _{n=1}^n\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}$

$S_n=\sum _{n=1}^n(\frac{1}{2n-1}-\frac{1}{2n+1})$

On putting the values of n

$S_n=\frac{1}{1}-\frac{1}{3}$

$+\frac{1}{3}-\frac{1}{5}$

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$+\frac{1}{2n-3}-\frac{1}{2n-1}$

$+\frac{1}{2n-1}-\frac{1}{2n+1}$

In this way all the middle terms will cancel each other and the resulting sum will be

$S_n=1-\frac{1}{2n+1}$

$S_n=\frac{2n+1-1}{2n+1}$

$S_n=\frac{2n}{2n+1}$.