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Question

If |x| < 1 then the coefficient of xn in the expansion

of (1+x+x2+.......)2 will be


A

1

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B

n

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C

n+1

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D

None of these

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Solution

The correct option is C

n+1


Given expression (1+x+x2+.......)2

= [(1x)1]2 = (1x)2

=(1+2x+3x2+4x3+........+(n1)xn+nxn1+......)

Therefore coefficient of xn is (n + 1).


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