Question

$1+3+7+13+...100terms=$

• A. $\frac{1010000}{2}$
• B. $\frac{1000200}{3}$
• C. $\frac{1015050}{3}$
• D. $\frac{1051050}{3}$

Answer 1

The correct option is B

$\frac{1000200}{3}$

Explanation for the correct option:

Step 1. Find the $nth$ of the series

Let $T_n$ is the $nth$ term of given series

$S=1+3+7+15+31+......+T_n$ …....(1)

$\Rightarrow S=1+3+7+15+......+T_{n-1}+T_n$….....(2)

Step 2. Subtracting equation (1) from (2), we get

$0=1+[2+4+6+8+16+.....(T_n-T_{n-1}\left)\right]-T_n$

$\Rightarrow$$T_n=1+\left[2+4+6+8+....+\left(n-1\right)\right]$

Here, $\left[2+4+6+8+....+\left(n-1\right)\right]$ is an AP

where, $a=2,d=2$

$\therefore T_n=1+\frac{\left(n-1\right)}{2}\left[2\left(2\right)+\left[\left(n-1\right)-1\right]\left(2\right)\right]$ $\left[\mathbf{\because }{\mathit{S}}_{\mathbf{n}}\mathbf{=}\mathbf{}\frac{\mathbf{n}}{\mathbf{2}}\left[2a+\left(n-1\right)d\right]\right]$

$=1+\left(\frac{(n-1)}{2}\right)\left(4+(n-2)2\right)$

$=1+(n-1)n$

$=n^2-n+1$

Step 3. Find the sum of given series:

$\begin{array}{rcl}S& =& \sum T_n\\ & =& \sum \left(n^2-n+1\right)\\ & =& \sum _{}{n}^2-\sum _{}n+\sum _{}1\end{array}$

$\begin{array}{rcl}& =& \frac{n(n+1)(2n+1}{6}-\frac{n(n+1)}{2}+1\end{array}$ $\left[\mathbf{\because }\mathbf{\sum }_{}{\mathit{n}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{;}\mathbf{}\mathbf{\sum }_{}\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{2}}\right]$

$$\begin{gathered} =\frac{n}{6}\left[\left(n+1\right)\left(2n+1\right)-3\left(n+1\right)+6\right] \\ =\frac{n}{6}\left[\left(n+1\right)\left(2n+1-3\right)+6\right] \\ =\frac{n}{6}\left[\left(n+1\right)\left(2n-2\right)+6\right] \\ =\frac{n}{3}\left[\left(n+1\right)\left(n-1\right)+3\right] \\ =\frac{n}{3}\left[n^2-1+3\right] \end{gathered}$$

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

Put $n=100$:

$\begin{array}{rcl}S& =& \frac{100\left({100}^2+2\right)}{3}\\ & =& \frac{100\left(10000+2\right)}{3}\end{array}$

$=\frac{\mathbf{1000200}}{\mathbf{3}}$

Hence, Option ‘B’ is Correct.