Sequences And Series Ques 12
- The sum of series $\frac{3 \times 1^3}{1^2}+\frac{5 \times\left(1^3+2^3\right)}{1^2+2^2}$ $+\frac{7 \times\left(1^3+2^3+3^3\right)}{1^2+2^2+3^2}+\ldots \ldots \ldots .+$ upto $10$ $th$ term, is
(2019 Main, 10 April I)
(a) $680$
(b) $600$
(c) $ 660$
(d) $ 620$
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Answer:
Correct Answer: 12.(c)
Solution: (c) Given series is
$ \frac{3 \times 1^3}{1^2}+\frac{5 \times\left(1^3+2^3\right)}{1^2+2^2}+\frac{7 \times\left(1^3+2^3+3^3\right)}{1^2+2^2+3^3}+\ldots $
So, $nth$ term
$ \begin{aligned} & T_n=\frac{(3+(n-1) 2)\left(1^3+2^3+3^3 \ldots+n^3\right)}{1^2+2^2+3^2+\ldots+n^2} \\ &=\frac{(2 n+1) \times\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)(2 n+1)}{6}} \\ & {\left[\because \sum_{r=1}^n r^3=\left[\frac{n(n+1)}{2}\right]^2 \text { and } \sum_{r=1}^n r^2=\frac{n(n+1)(2 n+1)}{6}\right] } \end{aligned} $
So, $\quad T_n=\frac{3 n(n+1)}{2}=\frac{3}{2}\left(n^2+n\right)$
Now, sum of the given series upto $n$ terms
$ \begin{aligned} S_n & =\Sigma T_n=\frac{3}{2}\left[\Sigma n^2+\Sigma n\right] \\ & =\frac{3}{2}\left[\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right] \\ \therefore \quad S_{10} & =\frac{3}{2}\left[\frac{10 \times 11 \times 21}{6}+\frac{10 \times 11}{2}\right] \\ & =\frac{3}{2}[(5 \times 11 \times 7)+(5 \times 11)] \\ & =\frac{3}{2} \times 55(7+1)=\frac{3}{2} \times 55 \times 8=3 \times 55 \times 4 \\ & =12 \times 55=660 \end{aligned} $
BETA
