Question

The sum of $(1+x)+(1+x+x^2)+(1+x+x^2+x^3)+\dots$ upto $n$ terms is

• A. $\frac{1}{1-x}[n-\frac{x^2(1-x^n)}{1-x}]$
• B. $\frac{1}{1-x}[n-\frac{x^3(1-x^n)}{1-x}]$
• C. $\frac{1}{1-x}[n-\frac{x(1-x^n)}{1-x}]$
• D. $\frac{1}{1-x}[n-\frac{2x(1-x^n)}{1-x}]$

Answer 1

The correct option is A $\frac{1}{1-x}[n-\frac{x^2(1-x^n)}{1-x}]$

$r^{\text{th}}$ term is given by
$T_r=1+x+x^2+\cdots +x^r=\frac{1-x^{r+1}}{1-x}$

Now, $\sum _{r=1}^n{T}_r$
$=\sum _{r=1}^n\frac{1-x^{r+1}}{1-x}$
$=\frac{1}{1-x}\sum _{r=1}^n(1-x^{r+1})$
$=\frac{1}{1-x}[n-\sum _{r=1}^n{x}^{r+1}]$
$=\frac{1}{1-x}[n-\frac{x^2(1-x^n)}{1-x}]$