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?

\frac{a}{1 - a} + \frac{a b}{1 - a b}

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\frac{a^{2}}{1 - a^{2}} + \frac{a b}{1 - a b}

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\frac{b}{1 - b} + \frac{a}{1 - a}

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\frac{b^{2}}{1 - b^{2}} + \frac{a b}{1 - a b}

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Solution

The correct option is B $\frac{a^{2}}{1 - a^{2}} + \frac{a b}{1 - a b}$
We have, $| a | < 1, | b | < 1$
$∴ | a b | = | a | | b | < 1$
Now, $a (a + b) + a^{2} (a^{2} + b^{2}) + a^{3} (a^{3} + b^{3}) + . . . . .$
$= [(a^{2} + a^{4} + a^{6} + . . . .)] + [{a b + (a b)^{2} + (a b)^{3} + . . . . .}]$
$= \frac{a^{2}}{1 - a^{2}} + \frac{a b}{1 - a b}$ ..... [Sum of infinite terms in G.P.]

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If |a|<1 and |b|<1, then the sum of the series a(a+b)+a2(a2+b2)+a3(a3+b3)+.... is?

The correct option is B a21−a2+ab1−abWe have, |a|<1,|b|<1∴|ab|=|a||b|<1Now, a(a+b)+a2(a2+b2)+a3(a3+b3)+.....=[(a2+a4+a6+....)]+[{ab+(ab)2+(ab)3+.....}]=a21−a2+ab1−ab ..... [Sum of infinite terms in G.P.]

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?