Question

If $a,x$ are real numbers and $|a|<1,|x|<1$, then $1+(1+a)x+(1+a+a^2)x^2+....\infty$ is equal to

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

Answer 1

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

Let $S=1+(1+a)x+(1+a+a^2)x^2+(1+a+a^2+a^3)x^3+....\infty$ ...$\left(1\right)$

Multiply above expression $\left(1\right)$ both sides by $ax$

$\Rightarrow axS=ax+(a+a^2)x^2+(a+a^2+a^3)x^3+......\infty$ ...$\left(2\right)$

Now subtract $\left(2\right)$ from $\left(1\right)$

$S-axS=1+x+x^2+x^3+.........\infty$

$\Rightarrow S(1-ax)=\frac{1}{1-x}$, sum of infinite G.P.

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