If the coefficient of x^3 in the expansion of _ r=0 ^ n (1+2x)^ n-r (1+x)^r is 4950 , then the value of n is:

Step-by-step solution

The given series is a geometric progression with the first term a = (1+2x)^n and the common ratio R = 1+x 1+2x . The number of terms in the series is n+1 . The sum of the series is given by S = a 1 - R^ n+1 1 - R : S = (1+2x)^n 1 - ( 1+x 1+2x )^ n+1 1 - 1+x 1+2x S = (1+2x)^n (1+2x)^ n+1 - (1+x)^ n+1 (1+2x)^ n+1 (1+2x) - (1+x) 1+2x S = (1+2x)^ n+1 - (1+x)^ n+1 x We are given that the coefficient of x^3 in S is 4950 . This is equivalent to saying that the coefficient of x^4 in the numerator (1+2x)^ n+1 - (1+x)^ n+1 i