If $| a | < 1$ and $| b | < 1$ then $s = 1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + . . . i s - .$

\frac{1}{(1 - b) (1 - a b)}

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

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

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\frac{1}{(1 + b) (1 + a b)}

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Solution

The correct option is D $\frac{1}{(1 - b) (1 - a b)}$
$S = (1 + b + b^{2} + b^{3} + . . . .) + a b (1 + b + b^{2} + . . . .) + a^{2} b^{2} (1 + b + b^{2} + . . .) + . . .$
Since using sum to infinity for G.P we get
$S = \frac{1}{1 - b} + \frac{a b}{1 - b} + \frac{a^{2} b^{2}}{1 - b} + . . . . S = \frac{1}{1 - b} (1 + a b + a^{2} b^{2} + . . . .)$
Using G.P for sum to infinity, we get
$S = \frac{1}{1 - b} \times \frac{1}{1 - a b} = \frac{1}{(1 - b) (1 - a b)}$

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