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

The correct option is D

$248$

Determine the value of $\lambda$

$A$ be the sum of the first $20$ terms

$$\begin{gathered} A=1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+...+2.{\left(20\right)}^2 \\ =\left(1^2+3^2+...+{19}^2\right)+\left(2.2^2+2.4^2+...+2.{\left(20\right)}^2\right)\left[\text{seperate odd and even term}\right] \\ =\left(1^2+3^2+...+{19}^2\right)+2.\left(2^2+4^2+...+{\left(20\right)}^2\right) \\ =\left(1^2+3^2+...+{19}^2\right)+\left(2^2+4^2+...+{\left(20\right)}^2\right)+\left(2^2+4^2+...+{\left(20\right)}^2\right) \\ =\left(1^2+2^2+3^2+...+{\left(20\right)}^2\right)+\left(2^2.1+2^2.2^2+...+2^2.{\left(10\right)}^2\right) \\ =\left(1^2+2^2+3^2+...+{\left(20\right)}^2\right)+2^2\left(1^2+2^2+3^2+...+{\left(10\right)}^2\right) \\ =\frac{20\times 21\times 41}{6}+4\left[\frac{10\times 11\times 21}{6}\right]\left[\because 1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\right] \\ =\frac{17220}{6}+\frac{4\times 2310}{6} \\ =4410 \end{gathered}$$

$B$ is the sum of first $40$ terms.

$$\begin{gathered} B=1^2+2.2^2+3^2+2.4^2+5^2+2.6^2+...+2.{\left(40\right)}^2 \\ =\left(1^2+3^2+...+{39}^2\right)+\left(2.2^2+2.4^2+...+2.{\left(40\right)}^2\right)\left[\text{seperate odd and even term}\right] \\ =\left(1^2+3^2+...+{39}^2\right)+2.\left(2^2+4^2+...+{\left(40\right)}^2\right) \\ =\left(1^2+3^2+...+{39}^2\right)+\left(2^2+4^2+...+{\left(40\right)}^2\right)+\left(2^2+4^2+...+{\left(40\right)}^2\right) \\ =\left(1^2+2^2+3^2+...+{\left(40\right)}^2\right)+\left(2^2.1+2^2.2^2+...+2^2.{\left(20\right)}^2\right) \\ =\left(1^2+2^2+3^2+...+{\left(40\right)}^2\right)+2^2\left(1^2+2^2+3^2+...+{\left(20\right)}^2\right) \\ =\frac{40\times 41\times 81}{6}+4\left[\frac{20\times 21\times 41}{6}\right]\left[\because 1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\right] \\ =\frac{132840}{6}+\frac{4\times 17220}{6} \\ =33620 \end{gathered}$$

We know that, $B-2A=100\lambda$

therefore, substituting the values we have,

$$\begin{gathered} \Rightarrow 33620-2\left(4410\right)=100\lambda \\ \Rightarrow 24800=100\lambda \\ \Rightarrow \lambda =248 \end{gathered}$$

Hence, option (D) is the correct answer