$\text{ Let }\mathrm{A}\text{ be the sum of the first }20\text{ terms and ' }\mathrm{B}\text{ ' be the sum of the first 40- }$$\text{ terms of the series }{1}^2+{2.2}^2+3^2+{2.4}^2+5^2+{2.6}^2+..\text{ If }\mathrm{B}-2\mathrm{A}=100\lambda \text{ then }\lambda \text{ is }$

$\text{B-2A=}\sum _{\text{n=1}}^{\text{40}}{\text{a}}_{\text{n}}\text{-2}\sum _{\text{n=1}}^{\text{20}}{\text{a}}_{\text{n}}\text{=}\sum _{\text{n=21}}^{\text{40}}{\text{a}}_{\text{n}}\text{-}\sum _{n=1}^{20}{\text{a}}_{\text{n}}$

$\text{B-2A=}\left({21}^2+{2.22}^2+{23}^2+\mathrm{...}+{40}^2\right)-\left(1^2+{2.2}^2+3^2+{2.4}^2+\mathrm{...}+{20}^2\right)$

$\begin{array}{l}=20(22+2.24+26+2.28+..+60)\\ =20\{\underset{20 terms}{\underbrace{(22+24+26+..+60)}}+\underset{10 terms}{\underbrace{(24+28+..+60)}}\}\\ =20\left\{\frac{20}{2}(22+60)+\frac{10}{2}(24+60)\right\}\\ =10(20.82+10.84)\\ =10(1640+840)\\ =100(164+84)=100\times 248\\ \to \lambda =248\end{array}$