Question

Let $S_1,S_2,S_3,S_4,S_5,\dots S_n$ are the sums of infinite geometric series whose first terms are $1,2,3,4,5,\dots ,n$ and whose common ratios are $\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}\dots ,\frac{1}{n+1}$ respectively. If $S_1^2+S_3^2+S_5^2+S_7^2+\dots +S_{99}^2=100k$, then find the value of $k$.

Answer 1

Answer: 1717

We have $S_1=\frac{1}{1-\frac{1}{2}}=2$,
$S_3=\frac{3}{1-\frac{1}{4}}=4$,
$S_5=\frac{5}{1-\frac{1}{6}}=6$
$\Rightarrow S_n=n+1,\forall n\in N$
Now $S_1^2+S_3^2+S_5^2+S_7^2+\dots +S_{99}^2=2^2+4^2+6^2+\dots$ $100^2$
$=2^2\left(1^2+2^2+3^2+4^2\dots 50^2\right)$
$=\left(2^2\right)\left(\frac{(50)(51)(101)}{6}\right)=17\times 10100$
$=(1717\times 100)=100k$
$\Rightarrow k=1717$