Question

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

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

Answer 1

Answer: (D)

$248$

$B=1^2+{2.2}^2+3^2+{2.4}^2+....+{2.40}^2$

$A=1^2+{2.2}^2+3^2+{2.4}^2+....+{2.20}^2$

$B=1^2+3^3+5^2+...+{39}^2+2[2^2+4^2+...+{40}^2]$

$A=1^2+3^2+5^2+...+{19}^2+2[2^2+4^2+...+{20}^2]$

Sum of square of first n odd natural number

$=\frac{n(2n+1)(2n-1)}{3}$

sum of square of first n even natural number

$=\frac{2n(n+1)(2n+1)}{3}$

$B={\left[\frac{n(2n+1)(2n-1)}{3}\right]}_{n=20}+2{\left[\frac{2n(n+1)(2n+1)}{3}\right]}_{n=20}$

$A={\left[\frac{n(2n+1)(2n-1)}{3}\right]}_{n=10}+2{\left[\frac{2n(n+1)2n+1)}{3}\right]}_{n=10}$

$B=\frac{20\left(41\right)\left(39\right)}{3}+2\left[\frac{\left(400\right(21\left)\right(41)}{3}\right]$

$B=\frac{100860}{3}$

$A=\frac{10\left(21\right)\left(19\right)}{3}+2\left[\frac{20\left(11\right)\left(21\right)}{3}\right]$

$A=\frac{13230}{3}$

$B-2A=\left[\frac{100860-26460}{3}\right]=\frac{74400}{3}$

$B-2A=24800=100\lambda$

$\lambda =248$