Proving the polynomial formula by induction

[quasar_4]

Homework Statement

(a1 + ... + ak)^n = n! $$\Sigma$$ (a^p)/p! = n! $$\Sigma$$ (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
formula).

Homework Equations

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?

 

[mighty2000]

Dear forum post author,

Thank you for your question. It seems like you have a good understanding of the induction process, but are unsure of how to apply it in this particular case. Let me try to guide you through the steps of the proof.

First, let's review the binomial formula for k = 2:

(a1 + a2)^n = n! * (a1^n + a1^(n-1)a2 + a1^(n-2)a2^2 + ... + a1a2^(n-1) + a2^n)

Now, to start the induction, we want to assume that the formula is true for k = m. In other words, we will assume that:

(a1 + a2 + ... + am)^n = n! * (a1^n + a1^(n-1)a2 + a1^(n-2)a2^2 + ... + a1a2^(n-1) + am^n)

Next, we want to show that this implies that the formula is also true for k = m+1. In other words, we want to show that:

(a1 + a2 + ... + am + a(m+1))^n = n! * (a1^n + a1^(n-1)a2 + a1^(n-2)a2^2 + ... + a1a2^(n-1) + a(m+1)^n)

To do this, we can use the binomial formula for k = m+1:

(a1 + a2 + ... + am + a(m+1))^n = n! * (a1^n + a1^(n-1)a2 + a1^(n-2)a2^2 + ... + a1a2^(n-1) + a(m+1)^n)

= n! * (a1^n + a1^(n-1)a2 + a1^(n-2)a2^2 + ... + a1a2^(n-1) + am^(n+1) + a(m+1)^n)

= n! * (a1^n + a1^(n-1)a2 + a1^(n-2)a2^2 + ... + a1a2^(n-1) + am^(n+1) + (a1+ a2 + ... +am)a(m+1)^n