Question

If $S_1,S_2,S_3,...S_n$ are sums of infinite geometric series whose first terms are $1,2,\mathrm{3..},n$ and whose common ratios are $\frac{1}{2},\frac{1}{3},\frac{1}{4},.....\frac{1}{n+1}$ respectively, then find the value of $S_1^2+S_2^2+S_3^2+.....+S_{2n-1}^2$

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

Answer 1

The correct option is B $\left[\frac{(n+1)(n+2)(2n+3)}{6}-1\right]+S_{n+1}^2+S_{n+2}^2+....+S_{2n-1}^2$

$S_1,S_2,S_3,...S_n$ are sums of infinite geometric series whose first terms are $1,2,\mathrm{3..},n$ and whose common ratios are $\frac{1}{2},\frac{1}{3},\frac{1}{4},.....\frac{1}{n+1}$ respectively
$S_1=\frac{1}{(1-\frac{1}{2})}=2$ ($\because S_{\infty }=\frac{a}{1-r}$)
$S_2=\frac{2}{(1-\frac{1}{3})}=3$, $S_3=\frac{3}{(1-\frac{1}{4})}=4$ .... $S_n=\frac{n}{(1-\frac{1}{n+1})}=n+1$
Therefore,
$S_1^2+S_2^2+S_3^2+.....+S_{2n-1}^2$
$=2^2+3^2+4^2+\cdots +(n+1{)}^2+S_{n+1}^2+S_{n+2}^2+....+S_{2n-1}^2$
$=\left[\frac{(n+1)(n+2)(2n+3)}{6}-1\right]+S_{n+1}^2+S_{n+2}^2+....+S_{2n-1}^2$ ($\because \sum n^2=\frac{n(n+1)(2n+1)}{6}$)
Hence, option B.