Question

The sum of the series $1+3x+5x^2+7x^3+\dots$ upto $n$ terms is

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

Answer 1

The correct option is B $\frac{1}{1-x}+2x\frac{(1-x^{n-1})}{(1-x{)}^2}-\frac{(2n-1)x^n}{(1-x)}$

$1+3x+5x^2+7x^2+\dots$
The given series is an AGP with $a=1,d=2,r=x,$ then
$S_n=\frac{a}{1-r}+dr\frac{(1-r^{n-1})}{(1-r{)}^2}-\frac{[a+(n-1\left)d\right]r^n}{1-r}$
$\Rightarrow S_n=\frac{1}{1-x}+2x\frac{(1-x^{n-1})}{(1-x{)}^2}-\frac{(2n-1)x^n}{1-x}$


Alternate Solution:
Let $S_n$ denote the sum of $n$ terms of the given series. Then,
$S_n=1+3x+5x^2+7x^3+\dots +(2n-3)x^{n-2}+(2n-1)x^{n-1}...\left(1\right)$
$x{S}_n=x+3x^2+5x^3+\dots +(2n-3)x^{n-1}+(2n-1)x^n...\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right),$ we have
$S_n-x{S}_n=1+[2x+2x^2+2x^3+\dots \dots +2x^{n-1}]-(2n-1)x^n$
$\Rightarrow S_n(1-x)=1+2x\left(\frac{1-x^{n-1}}{1-x}\right)-(2n-1)x^n$
$\Rightarrow S_n=\frac{1}{1-x}+2x\frac{(1-x^{n-1})}{(1-x{)}^2}-\frac{(2n-1)x^n}{1-x}$