Sequences And Series Ques 1
- The sum of series $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\ldots$ $+\frac{1^3+2^3+3^3+\ldots+15^3}{1+2+3+\ldots+15}-\frac{1}{2}(1+2+3+\ldots+15)$ is equal to
(2019 Main, 10 April II)
(a) $620$
(b) $660$
(c) $1240$
(d) $1860$
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Answer:
Correct Answer: 1.(a)
Solution: (a) Given series,
$ \begin{aligned} S & =1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\ldots+ \\ & \frac{1^3+2^3+3^3+\ldots+15^3}{1+2+3+\ldots+15}-\frac{1}{2}(1+2+3+\ldots+15) \\ & =S_1-S_2 \text { (let) } \end{aligned} $
where,
$ \begin{aligned} & S_1=1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\ldots+ \\ & \frac{1^3+2^3+8^3+\ldots+15^3}{1+2+3+\ldots+15} \\ & =\sum_{n=1}^{15} \frac{1^3+2^3+\ldots+n^3}{1+2+\ldots+n}=\sum_{n=1}^{15} \frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)}{2}} \\ & {\left[\because \sum_{r=1}^n r^3=\left(\frac{n(n+1)}{2}\right)^2 \text { and } \sum_{r=1}^n r=\frac{n(n+1)}{2}\right]} \\ & =\sum_{n=1}^{15} \frac{n(n+1)}{2}=\frac{1}{2} \sum_{n=1}^{15}\left(n^2+n\right) \\ & =\frac{1}{2}\left[\frac{15 \times 16 \times 31}{6}+\frac{15 \times 16}{2}\right] \\ & {\left[\because \sum_{r=1}^n r^2=\frac{n(n+1)(2 n+1)}{6}\right]} \\ & =\frac{1}{2}[(5 \times 8 \times 31)+(15 \times 8)] \\ & =(5 \times 4 \times 31)+(15 \times 4) \\ & =620+60=680 \\ & \text { and } S_2=\frac{1}{2}(1+2+3+\ldots+15) \\ & =\frac{1}{2} \times \frac{15 \times 16}{2}=60 \\ \end{aligned} $
Therefore, $S=S_1-S_2=680-60=620$.
BETA
