Question

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

• A. $5$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is A $5$

Given series is $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+...$
Here, $t_n=\frac{2n+1}{1^2+2^2+3^2+....+n^2}$
$\Rightarrow t_n=\frac{2n+1}{\frac{n(n+1)(2n+1)}{6}}$
$\Rightarrow t_n=\frac{6}{n(n+1)}$
$\Rightarrow S_n=6\sum \frac{1}{n(n+1)}$
$\Rightarrow S_n=6\sum (\frac{1}{n}-\frac{1}{n+1})$
So, $S_5=6{\sum }_{n=1}^5(\frac{1}{n}-\frac{1}{n+1})$
$\Rightarrow S_5=6(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6})$
$\Rightarrow S_5=6(1-\frac{1}{6})=5$