Question

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series
$1^2+2\cdot 2^2+3^2+2\cdot 4^2+5^2+2\cdot 6^2+\dots .$
If $B-2A=100\lambda$, then $\lambda$ is equal to

• A. $496$
• B. $232$
• C. $248$
• D. $464$

Answer 1

The correct option is C $248$

Since $A=$ Sum of the first $20$ terms
$\therefore A=(1^2+2^2+3^2+\dots +{20}^2)+(2^2+4^2+\dots +{20}^2)$
$\Rightarrow A=(1^2+2^2+3^2+\dots +{20}^2)+2^2(1^2+2^2+3^2+\dots +{10}^2)$
$\Rightarrow A=\frac{20\times 21\times 41}{6}+4\times \frac{10\times 11\times 21}{6}$
$\Rightarrow A=70\times 41+70\times 22$
$\Rightarrow A=4410$

And $B=(1^2+2^2+3^2+\dots +{40}^2)+(2^2+4^2+\dots +{40}^2)$
$\Rightarrow B=(1^2+2^2+3^2+\dots +{40}^2)+4(1^2+2^2+3^2+\dots +{20}^2)$
$\Rightarrow B=\frac{40\times 41\times 81}{6}+\frac{4\times 20\times 21\times 41}{6}$
$\Rightarrow B=22140+11480$
$\Rightarrow B=33620$

Now, $B-2A=33620-8820=24800$
$\therefore \lambda =248$