Question

If $S_1,S_2,S_3\dots \dots ,S_n,\dots$ are the sums of infinite geometric series whose first terms are $1,2,3,\dots \dots n,\dots \dots$ and whose common ratios are $\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots \dots \frac{1}{n+1}\dots \dots$ respectively, then the value of $\sum _{r=1}^{2n-1}{S}_r^2$ is

• A. $\frac{n(n+1)(2n+1)}{2}-1$
• B. $\frac{n(n+1)(n+2)}{2}-1$
• C. $\frac{n(2n+1)(4n+1)}{3}-1$
• D. $\frac{n(2n+1)(4n+1)}{3(n+1)}-1$

Answer 1

The correct option is C $\frac{n(2n+1)(4n+1)}{3}-1$

$r^{th}$ series will have $a=r$ and common ratio is $\frac{1}{r+1}$
$\therefore S_r=\frac{r}{1-\frac{1}{r+1}}=\frac{r(r+1)}{r}=r+1\therefore S_r^2=(r+1{)}^2$
$\therefore \sum _{r=1}^{2n-1}{S}_r^2=\sum _{r=1}^{2n-1}(r+1{)}^2=2^2+3^2+4^2+\dots \dots +(2n-1{)}^2+(2n{)}^2=1^2+2^2+3^2+\dots \dots +(2n{)}^2-1$
$=$ sum of the square of the first $\left(2n\right)$ natural numbers $=\frac{\left(2n\right)(2n+1)(4n+1)}{6}-1=\frac{n(2n+1)(4n+1)}{3}-1$