Question

The sum
$1+\left(1+x\right)+\left(1+x+x^2\right)+\left(1+x+x^2+x^3\right)+\dots n$ terms equals

• A. $\frac{1-x^n}{1-x}$
• B. $\frac{x\left(1-x^n\right)}{1-x}$
• C. $\frac{n\left(1-x\right)-x\left(1-x^n\right)}{{\left(1-x\right)}^2}$
• D. None of these

Answer 1

The correct option is C $\frac{n\left(1-x\right)-x\left(1-x^n\right)}{{\left(1-x\right)}^2}$

Let

$S_n=1+(1+x)+(1+x+x^2)+.......n$ terms

$x{S}_n=x+(x+x^2)+(x+x^2+x^3)+.....n$terms

Subtracting above series

$(1-x)S_n=n-(x+x^2+....+x^n)$

$(1-x)S_n=n-\frac{x(x^n-1)}{x-1}$.......As the series are in GP

$⟹S_n=\frac{n}{1-x}-\frac{x(x^n-1)}{(x-1{)}^2}$

Hence $S_n=\frac{n(1-x)-x(x^n-1)}{(1-x{)}^2}$