Question

Find $n$th term and sum of $n$ terms of the series: $1+3+7+15+31+....$

• A. $n$th term $T_n=(2^n-1)$. Sum of $n$ terms$=2^n-1-n$
• B. $n$th term $T_n=(2^n-1)$. Sum of $n$ terms$=2(2^n-1)-n$
• C. $n$th term $T_n=(2^n-1)$. Sum of $n$ terms$=2(2^n-1)-2n$
• D. $n$th term $T_n=(2^n-1-n)$. Sum of $n$ terms$=2(2^n-1)-n-1$

Answer 1

The correct option is B $n$th term $T_n=(2^n-1)$. Sum of $n$ terms$=2(2^n-1)-n$

Let $n$th term and sum of $n$ terms of series be denoted by $T_n$ and $S_n$ respectively.
$\therefore S_n=1+3+7+15+31+.....+T_{n-1}+T_n\cdots \left(1\right)$
$S_n=(1+3+7+15+......+T_{n-1})+T_n\cdots \left(2\right)$
Subtracting $\left(2\right)$ from $\left(1\right)$, we get

$0=(1+(3-1)+(7-3)+(15-7)+(31-15)+......+(T_n-T_{n-1})+T_n)$

$0=1+2+4+8+16+......+(T_n-T_{n-1})+T_n$
$\Rightarrow T_n=1+2+4+8+16+.....n$ terms $T_n=(2^n-1)$
$\therefore$ Sum of $n$ terms
$S_n=\sum T_n$
$=\sum (2^n-1)$
$=\sum 2^n-\sum 1$
$=(2+2^2+2^3+.....+2^n)-n$
$=2.\frac{(2^n-1)}{2-1}-n$
$=2(2^n-1)-n$