Question

If $|a|<1$ and $|b|<1$, then the sum of the series $a(a+b)+a^2(a^2+b^2)+a^3(a^3+b^3)+....$ is?

• A. $\frac{a}{1-a}+\frac{ab}{1-ab}$
• B. $\frac{a^2}{1-a^2}+\frac{ab}{1-ab}$
• C. $\frac{b}{1-b}+\frac{a}{1-a}$
• D. $\frac{b^2}{1-b^2}+\frac{ab}{1-ab}$

Answer 1

The correct option is B $\frac{a^2}{1-a^2}+\frac{ab}{1-ab}$

We have, $|a|<1,|b|<1$
$\therefore |ab|=|a||b|<1$
Now, $a(a+b)+a^2(a^2+b^2)+a^3(a^3+b^3)+.....$
$=\left[\right(a^2+a^4+a^6+....\left)\right]+\left[\right\{ab+(ab{)}^2+(ab{)}^3+.....\left\}\right]$
$=\frac{a^2}{1-a^2}+\frac{ab}{1-ab}$ ..... [Sum of infinite terms in G.P.]