Question

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

• A. 232
• B. 248
• C. 464
• D. 496

Answer 1

Answer: (B)

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