Sequences And Series Ques 34
- Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series
$ 1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+2 \cdot 6^2+\ldots $
If $B-2 A=100 $ $\lambda$, then $\lambda$ is equal to
(2018 Main)
(a) $232$
(b) $248$
(c) $464$
(d) $496$
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Answer:
Correct Answer: 34.(b)
Solution: (b) We have,
$ 1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+2 \cdot 6^2+\ldots $
$A=$ sum of first $20$ terms
$B=$ sum of first $40$ terms
$\therefore \mathrm{A}=1^2+2 \cdot 2^2+3^2+2 \cdot 4^2+5^2+2 \cdot 6^2+\ldots+2 \cdot 20^2$
$A=\left(1^2+2^2+3^2+\ldots+20^2\right)+\left(2^2+4^2+6^2+\ldots+20^2\right)$
$\left.A=1^2+2^2+3^2+\ldots+20^2\right)+4\left(1^2+2^2+3^2+\ldots+10^2\right)$
$\mathrm{A}=\frac{20 \times 21 \times 41}{6}+\frac{4 \times 10 \times 11 \times 21}{6}$
$A=\frac{20 \times 21}{6}(41+22)=\frac{20 \times 41 \times 63}{6}$
Similarly
$\left.B=\left(1^2+2^2+3^2+\ldots+40^2\right)+41^2+2^2+\ldots+20^2\right)$
$B=\frac{40 \times 41 \times 81}{6}+\frac{4 \times 20 \times 21 \times 41}{6}$
$B=\frac{40 \times 41}{6}(81+42)=\frac{40 \times 41 \times 123}{6}$
Now, $B-2 \mathrm{A}=100 \lambda$
$ \begin{aligned} & \therefore \frac{40 \times 41 \times 123}{6}-\frac{2 \times 20 \times 21 \times 63}{6}=100 \lambda \\ & \Rightarrow \frac{40}{6}(5043-1323)=100 \lambda \Rightarrow \frac{40}{6} \times 3720=100 \lambda \\ & \Rightarrow 40 \times 620=100 \lambda \Rightarrow \lambda=\frac{40 \times 620}{100}=248 \end{aligned} $
BETA
