Let C(m,r) be the coefficient of x^r in d(m) = (x + x^2 + .. + x^n)^m. Then C(m,r) is 0 whenever m>r.
Using recursion relations or induction, we come up for an expression for C(m,r) when n>=m>=r
C(m,r) = (r1)!/(m1)!(rm)!
So, the coefficient of x^r in (Sum over m) d(m) is given by
S(r) = (Sum of m from 1 to r) (1)^m (r1)!/(m1)!(rm)!
Now, look at the binomial expansion of G(y) = (1y)^(r1). This is given by
G(y) = (Sum of m from 1 to k) (1)^m y^m (r1)!/(m1)!(rm)!
Hence 0 = G(1) = S(r)
Edited on February 4, 2019, 9:31 am

Posted by FrankM
on 20190204 07:04:44 