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) = (r-1)!/(m-1)!(r-m)!
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 (r-1)!/(m-1)!(r-m)!
Now, look at the binomial expansion of G(y) = (1-y)^(r-1). This is given by
G(y) = (Sum of m from 1 to k) (-1)^m y^m (r-1)!/(m-1)!(r-m)!
Hence 0 = G(1) = S(r)
Edited on February 4, 2019, 9:31 am
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Posted by FrankM
on 2019-02-04 07:04:44 |
No Subject