The coefficient of x⁵⁰ in the binomial expansion of (1+x)¹⁰⁰⁰+x(1+x)⁹⁹⁹+x^2(1+x)⁹⁹⁸+ +x¹⁰⁰⁰ is:

Step-by-step solution

Let given expansion be aligned & S =(1+x)¹⁰⁰⁰+x(1+x)⁹⁹⁹+x^2 &(1+x)⁹⁹⁸+ + +x¹⁰⁰⁰ aligned Put 1+x=t S =t¹⁰⁰⁰+x t⁹⁹⁹+x^2(t)⁹⁹⁸+ +x¹⁰⁰⁰ This is a G.P with common ratio x t aligned S &= t¹⁰⁰⁰ [1- ( x t )¹⁰⁰¹ ] 1- x t &= (1+x)¹⁰⁰⁰ [1- ( x 1+x )¹⁰⁰¹ ] 1- x 1+x =& (1+x)¹⁰⁰¹ [(1+x)¹⁰⁰¹-x¹⁰⁰¹ ] (1+x)¹⁰⁰¹ aligned = [(1+x)¹⁰⁰¹-x¹⁰⁰¹ ] Now coeff of x⁵⁰ in above expansion is equal to coeff of x⁵⁰ in (1+x)¹⁰⁰¹ which is 1001 50 = (1001) ! 50 !(951) !