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Question

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

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


  1. None of these

  2. 1

  3. n

  4. n+1


Solution

The correct option is D

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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