Question

The sum of $1+\left(1+a\right)x+\left(1+a+a^2\right)x^2+....\infty ,0<a,x<1$ equals

• A. $\frac{1}{\left(1-x\right)\left(1-a\right)}$
• B. $\frac{1}{\left(1-a\right)\left(1-ax\right)}$
• C. $\frac{1}{\left(1-x\right)\left(1-ax\right)}$
• D. $\frac{1}{\left(1-x\right)\left(1+a\right)}$

Answer 1

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

$1+(1+a)x+(1+a+a^2)x^2+\dots \infty$

$Generaltermofseries=T_n=\frac{(1-a^{r+1})}{(1-a)}{x}^r$

$S_n={\sum }_{r=1}^{\infty }\frac{(1-a^{r+1})x^r}{(1-a)}$

$(1-a)S_n={\sum }_{r=1}^{\infty }{x}^r-a{\sum }_{r=1}^{\infty }{\left(ax\right)}^r$

$(1-a)S_n=\frac{1}{1-x}-\frac{a}{1-ax}$ ........... $\because G.P.$ series

$(1-a)S_n=\frac{1-ax-a+ax}{(1-x)(1-ax)}$

$(1-a)S_n=\frac{1-a}{(1-x)(1-ax)}$

$S_n=\frac{1}{(1-x)(1-ax)}$

B is correct