Sequences And Series Ques 4
- The sum of first $9$ terms of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\ldots$ is
(2015)
(a) $71$
(b) $96$
(c) $142$
(d) $192$
Show Answer
Answer:
Correct Answer: 4.(b)
Solution: (b) PLAN
Write the nth term of the given series and simplify it to get its lowest form. Then, apply, $\mathrm{S}_n=\sum \mathrm{T}_n$
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{\left\{\frac{n(n+1)}{2}\right\}^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}\left[\frac{10(10+1)(20+1)}{6}-1\right]=\frac{384}{4}=96 \end{aligned} $
BETA
