Question

The sum of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....$ upto $n$ terms, is

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

Answer 1

The correct option is C $\frac{6n}{n+1}$

The $n$th term in the given series can be written as,

$t_n=\frac{2n+1}{1^2+2^2+.....+n^2}$

or, $t_n=\frac{2n+1}{\frac{n(n+1)(2n+1)}{6}}$

or, $t_n=\frac{6}{n(n+1)}$

or, $t_n=\frac{-6}{(n+1)}+\frac{6}{n}$.

The sum of the series upto $n$ terms will be,

$\sum _{k=1}^{k=n}{t}_k$

$=\sum _{k=1}^{k=n}(\frac{6}{k}-\frac{6}{k+1})$

$=\frac{6}{1}-\frac{6}{2}+\frac{6}{2}-\frac{6}{3}+\frac{6}{3}..........-\frac{6}{n}+\frac{6}{n}-\frac{6}{n+1}.$

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

$=\frac{6n}{n+1}$.