(a1 + ... + ak)^n = n! [tex]\Sigma[/tex] (a^p)/p! = n! [tex]\Sigma[/tex] (a1^p1 *** ak^pk)/(p1!***pk!)
(the 1...k are subscripts, not coefficients)
Prove this formula through induction on k beginning with the case k = 2 (binomial
Just the above formula; the binomial formula is the case where k =2, which is where my induction should begin.
The Attempt at a Solution
Well, I think I can kind of do the k=2 step (with minor confusion about indices that I should be able to resolve). But what confuses me is the next part. I'm supposed to be doing the induction on k, so I guess I assume it's true for k = m, then show that it's true for k = m+1. But I'm confused about how to do this... if it's true for k = m then I have (a1+...+am)^ n = the formula above; how do I show anything for m+1? If I put in
(a1+...+a(m+1))^n, I can't just multiply things out... so I'm a bit confused. Any suggestions?