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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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xyx−y−1

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Correct option is D

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+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