Question

Find the sum of the n terms of the series $1+3+7+15+31+\cdots n$ terms.

• A. $2^{(n+1)}-3-n$
• B. $2^{(n+1)}-2+n$
• C. $2^{(n+1)}+3+n$
• D. $2^{(n+1)}-2-n$

Answer 1

The correct option is D

$2^{(n+1)}-2-n$

We see that the differences between successive a term is $2,4,8,16,\cdots$

This is a GP.
$S_n=1+3+7+15+31+\cdots T_{n-1}+T_n$
$\cdots \left(1\right)$
$S_n=1+3+7+15+\cdots T_{n-1}+T_n$
$\cdots \left(2\right)$

$[$Now, we write the $S_n$ in such a way that $1^{st}$ term of equation 2 comes under $2^{nd}$ term of equation 1]

Then, Subtracting both, we get,
$0=1+2+4+8+16+\cdots +(T_n+T_{n-1})-T_n$
$T_n=1+2+4+8+16+\cdots n$ terms
we get, $T_n$ = $\frac{1.(2^n-1)}{2-1}$
Hence sum of the $n$ terms
$S_n=\sum t_n={\sum }_{i=1}^n(2^i-1)$
$S_n=\sum 2^n-\sum 1=2^{(n+1)}-2-n$

Aliter:
${\S }_n=1+3+7+15+\cdots +n^{th}\text{term}=(2^1-1+2^2-1+2^3-1+\cdots +2^n-1=2^1+2^2+2^3+\cdots +2^n-n=2\times \frac{1-2^n}{1-2}-n=-2+2^{n+1}-n$