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Question

$$x= 1+a+a^2+....\infty , y= 1+b+b^2+....\infty ,\left | ab \right |<1,$$  then $$1+ab+a^2b^2+....\infty =$$


A
xyx+y1
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B
x+y1yx
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C
xyx+y+1
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D
x+y+1xy
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Solution

The correct option is D $$\frac{xy}{x+y-1}$$
$$ x\; =\; \frac{1}{1-a}$$ 
$$ y\; =\; \frac{1}{1-b} $$
i.e $$ a\;= 1\;-\frac{1}{x}$$ and $$ b\;= 1\;-\;\frac{1}{y} $$
$$ab\;=\;(1\;-\frac{1}{x}) (1\;-\;\frac{1}{y}) $$
$$\therefore 1+ab+{ab}^2 +........ \;= \frac{1}{1-ab}=\frac{xy}{x+y-1}$$

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