Sequences and Series 5 Question 21
12. If $S _1, S _2, S _3, \ldots, S _n$ are the sums of infinite geometric series, whose first terms are $1,2,3, \ldots, n$ and whose common ratios are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{n+1}$ respectively, then find the values of $S _1^{2}+S _2^{2}+S _3^{2}+\ldots+S _{2 n-1}^{2}$.
(1991, 4M)
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Solution:
- PLAN Write the $n$th term of the given series and simplify it to get its lowest form. Then, apply, $S _n=\sum T _n$
Given series is $\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$
Let $T _n$ be the $n$th term of the given series.
$$ \begin{aligned} \therefore T _n & =\frac{1^{3}+2^{3}+3^{3}+\ldots+n^{3}}{1+3+5+\ldots+\text { upto } n \text { terms }} \\ & =\frac{\frac{n(n+1)^{2}}{2}}{n^{2}}=\frac{(n+1)^{2}}{4} \\ S _9 & \left.=\sum _{n=1}^{9} \frac{(n+1)^{2}}{4}=\frac{1}{4}\left(2^{2}+3^{2}+\ldots+10^{2}\right)+1^{2}-1^{2}\right] \\ & =\frac{1}{4} \frac{10(10+1)(20+1)}{6}-1=\frac{384}{4}=96 \end{aligned} $$
