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Question

If |x|<1,then the coefficient of xn in the expansion of (1+x+x2+x3+....)2 is

A
n
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B
n1
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C
n+2
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D
n+1
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Solution

The correct option is D n+1
As 1+x+x2+x3+=(1x)1
(1+x+x2+.....)2=(1x)2
=1+2x+3x2+......
Therefore, coefficient of xn is n+1.

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