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

This type of question is based on the concept of sequences and series. We need to know the formula of the sum of first $n$ natural numbers squared to solve this problem, which is,
$\sum\limits_{i=1}^{n}{{{i}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
We need to find specific patterns in the given question and rearrange the terms in such a way that we can use this formula to simplify the problem and find the solution.

Complete step by step solution:
In the question, we are given that $A$ is the sum of the first $20$ terms and $B$ is the sum of the first $40$ terms of the series ${{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+.....$ And $B-2A=100\lambda$. We are asked to find the value of $\lambda$ .
First, let us focus on the L.H.S which is $B-2A$ . We need to find the values of $B$ and $A$ . we are given that $A$ is the sum of first $20$ terms of the series ${{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+.....$therefore, $A={{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+{{.....2.20}^{2}}$ . By inspecting the series, we can find that every odd term is of the form ${{a}_{n}}={{n}^{2}}$ , where $n$ is odd and the ${{n}^{th}}$ term of the series. Also, we can find that every even term is of the form ${{a}_{n}}=2.{{n}^{2}}$ , where $n$ is even and the ${{n}^{th}}$ term of the series. We can now write $A$ as,

& A=\left( {{1}^{2}}+{{3}^{2}}+....+{{19}^{2}} \right)+\left( {{2.2}^{2}}+{{2.4}^{2}}+......+{{2.20}^{2}} \right) \\\ & \Rightarrow A=\left( {{1}^{2}}+{{3}^{2}}+....+{{19}^{2}} \right)+2\left( {{2}^{2}}+{{4}^{2}}+......+{{20}^{2}} \right) \\\ & \Rightarrow A=\left( {{1}^{2}}+{{3}^{2}}+......+{{19}^{2}} \right)+\left( {{2}^{2}}+{{4}^{2}}+......+{{20}^{2}} \right)+\left( {{2}^{2}}+{{4}^{2}}+......+{{20}^{2}} \right) \\\ & \Rightarrow A=\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}......+{{19}^{2}}+{{20}^{2}} \right)+4\left( {{1}^{2}}+{{2}^{2}}+{{......10}^{2}} \right) \\\ \end{aligned}$$ Now, we can use the formula of the sum of first $n$ natural numbers squared, which is, $\sum\limits_{i=1}^{n}{{{i}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ Therefore, $\begin{aligned} & A=\left( sum\,of\,first\,\,20\,\,natural\,numbers\,squared \right) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,+4\left( sum\,of\,first\,\,10\,\,natural\,numbers\,squared \right) \\\ & \Rightarrow A=\dfrac{20\left( 20+1 \right)\left( 2.20+1 \right)}{6}+\dfrac{10\left( 10+1 \right)\left( 2.10+1 \right)}{6} \\\ & \Rightarrow A=\dfrac{20\left( 21 \right)\left( 41 \right)}{6}+\dfrac{10\left( 11 \right)\left( 21 \right)}{6} \\\ & \Rightarrow A=2870+1540 \\\ \end{aligned}$ $\begin{aligned} & \,\,\Rightarrow \,A=4410 \\\ & \Rightarrow 2A=8820......\left( i \right) \\\ \end{aligned}$ Similarly, $B$ is the sum of first $40$ terms of the series ${{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+.....$therefore, $A={{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+{{.....2.20}^{2}}$. We can now write $B$ as, $$\begin{aligned} & B=\left( {{1}^{2}}+{{3}^{2}}+....+{{39}^{2}} \right)+\left( {{2.2}^{2}}+{{2.4}^{2}}+......+{{2.40}^{2}} \right) \\\ & \Rightarrow B=\left( {{1}^{2}}+{{3}^{2}}+....+{{39}^{2}} \right)+2\left( {{2}^{2}}+{{4}^{2}}+......+{{40}^{2}} \right) \\\ & \Rightarrow B=\left( {{1}^{2}}+{{3}^{2}}+......+{{39}^{2}} \right)+\left( {{2}^{2}}+{{4}^{2}}+......+{{40}^{2}} \right)+\left( {{2}^{2}}+{{4}^{2}}+......+{{40}^{2}} \right) \\\ & \Rightarrow B=\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}......+{{39}^{2}}+{{40}^{2}} \right)+4\left( {{1}^{2}}+{{2}^{2}}+{{......20}^{2}} \right) \\\ \end{aligned}$$ Therefore, $\begin{aligned} & B=\left( sum\,of\,first\,\,40\,\,natural\,numbers\,squared \right) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,+4\left( sum\,of\,first\,\,20\,\,natural\,numbers\,squared \right) \\\ & \Rightarrow B=\dfrac{40\left( 40+1 \right)\left( 2.40+1 \right)}{6}+\dfrac{20\left( 20+1 \right)\left( 2.20+1 \right)}{6} \\\ & \Rightarrow B=\dfrac{40\left( 41 \right)\left( 81 \right)}{6}+\dfrac{20\left( 21 \right)\left( 41 \right)}{6} \\\ & \Rightarrow B=22140+11480 \\\ \end{aligned}$ $\,\,\,\Rightarrow B=33620......\left( ii \right)$ From equations $\left( i \right)$ and $\left( ii \right)$ , we get, $\begin{aligned} & B-2A=33620-8820 \\\ & B-2A=24800 \\\ & \Rightarrow B-2A=100\lambda \\\ & \Rightarrow 100\lambda =24800 \\\ \end{aligned}$ $\Rightarrow \lambda =248$ **So, the correct answer is “Option d”.** **Note:** While solving this problem, one has to be careful while grouping the terms as it is prone to calculation mistakes. Even if one term is not grouped correctly, the answer will vary by a large margin. Hence one has to group the terms of the series carefully. Also, the last term of a series is very significant and usually one tends to make errors in this. So, write the generalised series equation first and then find the last term using that. For example, we found that last term or the ${{20}^{th}}$ term of $A$ is ${{2.20}^{2}}$ because we knew that every even term is of the form ${{a}_{n}}=2.{{n}^{2}}$, where $n$ is odd and the ${{n}^{th}}$ term of the series. Hence, write the generalised term first then substitute the term which u want whether it is odd or even term.