Question

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series $1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + ….$ If $B - 2A = 100\lambda$, then $\lambda$ is equal to :

• A. 232
• B. 248
• C. 464
• D. 496

Answer 1

Answer: (B)

$A =1^{2}+2.2^{2}+3^{2}+\ldots .+2.20^{2} $
$ =\left(1^{2}+2^{2}+3^{2}+\ldots+20^{2}\right)+4\left(1^{2}+2^{2}+3^{2}+\ldots+10^{2}\right) $
$ =\frac{20 \times 21 \times 41}{6}+\frac{4 \times 10 \times 11 \times 21}{6} $
$=2870+1540=4410 $
$ B =1^{2}+2.2^{2}+3^{2}+\ldots .+2.40^{2} $
$=\left(1^{2}+2^{2}+3^{2}+\ldots .+40^{2}\right)+4\left(1^{2}+2^{2}+3^{2}+\ldots .+20^{2}\right)$
$=\frac{40 \times 41 \times 81}{6}+\frac{4 \times 20 \times 21 \times 41}{6}$
$=22140+11480=33620$
$ \Rightarrow B-2 A=33620-8820=24800$
$\Rightarrow 100 \lambda=24800 $
$ \lambda=248 $