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 $B-2A=100\lambda ,\text{ then }\lambda$  is equal to

We have, 

$1^2+2\cdot 2^2+3^2+2\cdot 4^2+5^2+2\cdot 6^2+\dots$

$A=$ sum of first 20 terms

B = sum of first 40 terms

$\begin{array}{l}\therefore A=1^2+2\cdot 2^2+3^2+2\cdot 4^2+5^2+2\cdot 6^2+\dots +2\cdot {20}^2\\ A=\left(1^2+2^2+3^2+\dots +{20}^2\right)+\left(2^2+4^2+6^2+\dots +{20}^2\right)\\ A=\left(1^2+2^2+3^2+\dots +{20}^2\right)+4\left(1^2+2^2+3^2+\dots +{10}^2\right)\\ A=\frac{20\times 21\times 41}{6}+\frac{4\times 10\times 11\times 21}{6}\\ A=\frac{20\times 21}{6}(41+22)\\ =\frac{20\times 41\times 63}{6}\end{array}$

Similarly,    

$\begin{array}{l}B=\left(1^2+2^2+3^2+\dots +{40}^2\right)+4\left(1^2+2^2+\dots +{20}^2\right)\\ B=\frac{40\times 41\times 81}{6}+\frac{4\times 20\times 21\times 41}{6}\\ B=\frac{40\times 41}{6}(81+42)=\frac{40\times 41\times 123}{6}\end{array}$

Now,  $B-2A=100\lambda$

$\begin{array}{l}\therefore \frac{40\times 41\times 123}{6}-\frac{2\times 20\times 21\times 63}{6}=100\lambda \\ \Rightarrow \frac{40}{6}(5043-1323)=100\lambda \\ \Rightarrow \frac{40}{6}\times 3720=100\lambda \\ \Rightarrow 40\times 620=100\lambda \\ \Rightarrow \lambda =\frac{40\times 620}{100}=248\end{array}$