If x=1+a+a2+…(|a|<1) and y=1+b+b2+…(|b|<1) then some of the series 1+ab+a2b2+… is

x=1+a+a2+⋯=11−ay=11−b⇒a=1−1/x,b=1−1/yNow 1+ab+a2b2+…=11−ab=11−(1−1/x)(1−1/y)=xyx+y−1

If x=1+a+a2+…(|a|<1) and y=1+b+b2+…(|b|<1) then some of the series 1+ab+a2b2+… is

x=1+a+a2+⋯=11−ay=11−b⇒a=1−1/x,b=1−1/yNow 1+ab+a2b2+…=11−ab=11−(1−1/x)(1−1/y)=xyx+y−1

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If $x = 1 + a + a^{2} + \dots (| a | < 1)$ and $y = 1 + b + b^{2}$ $+ \dots (| b | < 1)$ then some of the series $1 + a b + a^{2} b^{2} + \dots$ is

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$\frac{x y}{x + y - 1}$

$\frac{x y}{x - y + 1}$

$\frac{x y}{x + y + 1}$

$\frac{x y}{x - y - 1}$

answer is D.

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$\begin{array}{l} x = 1 + a + a^{2} + \dots = \frac{1}{1 - a} \\ y = \frac{1}{1 - b} \\ \Rightarrow a = 1 - 1 / x, b = 1 - 1 / y \end{array}$

Now $1 + a b + a^{2} b^{2} + \dots$

$\begin{array}{l} = \frac{1}{1 - a b} = \frac{1}{1 - (1 - 1 / x) (1 - 1 / y)} \\ = \frac{x y}{x + y - 1} \end{array}$

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If x=1+a+a2+…(|a|<1) and y=1+b+b2+…(|b|<1) then some of the series 1+ab+a2b2+… is

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If $x = 1 + a + a^{2} + \dots (| a | < 1)$ and $y = 1 + b + b^{2}$ $+ \dots (| b | < 1)$ then some of the series $1 + a b + a^{2} b^{2} + \dots$ is