If |x| < 1 then the coefficient of xn in the expansion
of (1+x+x2+.......)2 will be
1
n
n+1
None of these
Given expression (1+x+x2+.......)2
= [(1−x)−1]2 = (1−x)−2
=(1+2x+3x2+4x3+........+(n−1)xn+nxn−1+......)
Therefore coefficient of xn is (n + 1).
Coefficient of x100in1+(1+x)+(1+x)2+(1+x)3…(1+x)n is 201C101,(n>100) then n =
If |x| < 1, then in the expansion of
(1+2x+3x2+4x3+.......)12, the coefficient of xn is