Mathematics - Sequence & Series Question with Solution | TestHub

Given series, $1^2+{2·2}^2+3^2+{2·4}^2+5^2+{2·6}^2+\dots$

Here, $A=1^2+2·2^2+3^2+2·4^2+...20$ terms

$=\left(1^2+ 2^2+3^2+4^2+...20 \mathrm{terms}\right)+\left(2^2+4^2+ 6^2+...10 \mathrm{terms}\right)$

$=\frac{20·21·41}{6} +4\left(\frac{10·11·21}{6}\right)$ 

$\Rightarrow A=4410$

Similarly for $B=\left(1^2+2^2+3^3+...40 \mathrm{terms}\right)+\left(2^2+4^2+6^2+...20 \mathrm{terms}\right)$

$\Rightarrow B=\frac{40\times 41\times 81}{6}+\frac{4\times 20\times 21\times 41}{6}$

$\Rightarrow B=33620$

Now, $B-2A=24800$$=100\lambda$ (Given)

$\Rightarrow \lambda =248$