Question

$\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+...nterms=?$

• A. $\frac{6n^2}{(n+1)}$
• B. $\frac{6n}{(n+1)}$
• C. $\frac{6(2n-1)}{(n+1)}$
• D. $\frac{3(n^2+1)}{(n+1)}$

Answer 1

The correct option is B

$\frac{6n}{(n+1)}$

Explanation for the correct option:

Step 1. Find the $nth$ term of given series:

$\Rightarrow T_n=\frac{2n+1}{{\displaystyle \frac{n(n+1)(2n+1)}{6}}}=\frac{6}{n(n+1)}$

$\Rightarrow T_n=6\left[\frac{1}{n}-\frac{1}{n+1}\right]$

Step 2. Find the sum of $T_n$:

Therefore, $S_n=\sum _{}{T}_n=6\sum _{}\left(\frac{1}{n}-\frac{1}{n+1}\right)$

$\Rightarrow 6\sum _{}\frac{1}{n}-6\sum _{}\frac{1}{n+1}$

$\Rightarrow \frac{6n}{n}-\frac{6}{n+1}=6-\frac{6}{n+1}$

$\Rightarrow \frac{\mathbf{6}\mathbf{n}}{\mathbf{n}\mathbf{+}\mathbf{1}}$

Hence, Option ‘B’ is Correct.