Question

Let $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}, \ldots S_{n}$ are the sums of infinite geometric series whose first terms are $1,2,3,4,5, \ldots, n$ and whose common ratios are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6} \ldots, \frac{1}{n+1}$ respectively. If $S_{1}^{2}+S_{3}^{2}+S_{5}^{2}+S_{7}^{2}+\ldots+S_{99}^{2}=100 k$, then find the value of $k$.

Answer 1

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 $S_{1}^{2}+S_{3}^{2}+S_{5}^{2}+S_{7}^{2}+\ldots+S_{99}^{2}=2^{2}+4^{2}+6^{2}+\ldots$ $100^{2}$
$=2^{2}\left(1^{2}+2^{2}+3^{2}+4^{2} \ldots 50^{2}\right)$
$=\left(2^{2}\right)\left(\frac{(50)(51)(101)}{6}\right)=17 \times 10100$
$=(1717 \times 100)= 100 \,k$
$\Rightarrow k=1717$