Question

For first $n$ natural numbers we have the following results with usual notations $\sum _{r=1}^{n}r=\frac{n(n+1)}{2},\sum _{r=1}^n{r}^2=\frac{n(n+1)(2n+1)}{6},\sum _{r=1}^n{r}^3={\left(\sum _{r=1}^{n}r\right)}^2$ If $a_1{a}_2....a_n\in A.P$ then sum to $n$ terms of the sequence $\frac{1}{a_1{a}_2},\frac{1}{a_2{a}_3},...\frac{1}{a_{n-1}{a}_n}$ is equal to $\frac{n-1}{a_1{a}_n}$
and the sum to $n$ terms of a $G.P$ with first term '$a$' & common ratio '$r$' is given by $S_n=\frac{lr-a}{r-1}$ for $r\ne 1$ for $r=1$ sum to $n$ terms of same $G.P.$ is $n$ $a$, where the sum to infinite terms of$G.P.$ is the limiting value of
$\frac{lr-a}{r-1}$ when $n\to \infty ,\left|r\right|<l$ where $l$ is the last term of $G.P.$ On the basis of above data answer the following questionsThe sum to infinite terms of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+...\infty$ 15 equal to?

• A. 4
• B. 6
• C. 8
• D. None of these

Answer 1

The correct option is B 6

Let $S_n=\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+...+\frac{2n+1}{1^2+2^2+...+n^2}$
Now, $t_r=\frac{2r+1}{1^2+2^2+...+r^2}=\frac{6}{r(r+1)}$
Therefore, $S_n={\sum }_{r=1}^n{t}_r={\sum }_{r=1}^n\frac{6}{r(r+1)}=6{\sum }_{r=1}^n(\frac{1}{r}-\frac{1}{r+1})=\frac{6n}{n+1}$
$S_{\infty }={lim}_{n\to \infty }{S}_n=6{lim}_{n\to \infty }\frac{n}{n+1}=6{lim}_{n\to \infty }\frac{1}{1+\frac{1}{n}}=6$
Ans: B