Sequences And Series Ques 23
- The sum of the series $1+2 \times 3+3 \times 5+4 \times 7+\ldots$ upto $11$ $ th$ term is
(2019 Main, 9 April II)
(a) $915$
(b) $946$
(c) $916$
(d) $945$
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Answer:
Correct Answer: 23.(b)
Solution: (b) Given series is
$ 1+(2 \times 3)+(3 \times 5)+(4 \times 7)+\ldots \text { upto } $ $11$ terms.
Now, the $r$ th term of the series is $a_r=r(2 r-1)$
$\therefore \quad $ Sum of first $11$-terms is
$ \begin{aligned} S_{11} & =\sum_{r=1}^{11} r(2 r-1)=\sum_{r=1}^{11}\left(2 r^2-r\right)=2 \sum_{r=1}^{11} r^2-\sum_{r=1}^{11} r \\ & =\frac{11 \times(11+1)(2 \times 11+1)}{6}-\frac{11 \times(11+1)}{2} \\ & =\left(\because \sum_{r=1}^n r^2=\frac{n(n+1)(2 n+1)}{6} \text { and } \sum_{r=1}^n r=\frac{n(n+1)}{2}\right] \\ & =\left(\frac{11 \times 12 \times 23}{3}\right)-\left(\frac{11 \times 12}{2}\right) \end{aligned} $
$ \quad =(11 \times 4 \times 23)-(11 \times 6)=11(92-6)=11 \times 86=946 $
BETA
