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Question

If |x|<1,|y|<1 and xy, then the sum to infinity of the following series (x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+......

A
x+y+xy(1x)(1y)
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B
x+yxy(1x)(1y)
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C
x+y+xy(1+x)(1+y)
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D
x+yxy(1+x)(1+y)
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Solution

The correct option is B x+yxy(1x)(1y)
(x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+......

=1(xy){(x2y2)+(x3y3)+(x4y4)+....} =x21xy21yxy

=x2(1y)y2(1x)(1x)(1y)(xy)

=(x2y2)xy(xy)(1x)(1y)(xy)=((x+y)xy(xy))(1x)(1y)(xy)

=x+yxy(1x)(1y)

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