# Proof using the Binomial Theorem

1. Oct 6, 2011

### Karnage1993

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I am really stuck, I have no clue how to even begin. For part B I tried changing the RHS to factorials but I was left at a dead end there.

2. Oct 6, 2011

### Dick

They gave you a nice clue for the first part. Expand (1+x)^n and (1+x)^m and multiply them. Then do the equal expression (1+x)^(n+m). Equate equal powers of x.

3. Oct 6, 2011

### Karnage1993

Yes, I get two sums being multiplied together. That part I don't know how to do => simplifying sums multiplied with each other.

4. Oct 6, 2011

### Dick

Ok, then start simple. (1+x)^3 times (1+x)^3 is the same as (1+x)^6. Write the expansion of them in terms of binomial coefficients, not numbers. Now equate equal powers of x. It's a little tedious, but it's worth doing this if you really don't see what to do.

5. Oct 6, 2011

### Karnage1993

The coefficients would be 1 + 6 + 15 + 20 + 15 + 6 + 1. What do you mean by equate equal powers of x? I am not familiar with that term. Sorry, I'm only a first year student studying Spivak. :)

6. Oct 6, 2011

### Dick

(1+x)^3=C(3,0)+C(3,1)*x+C(3,2)*x^2+C(3,3)*x^3. I meant expand it like that. I know they are simple numbers but don't write 1, 6, 15 etc. Write C(6,0), C(6,1), C(6,2) etc. That's the only way you are going to see what's going on. And equating equal powers just means if 1+2x+3x^2=a+bx+cx^2 for all x then a=1, b=2 and c=3. That sort of thing.

7. Oct 6, 2011

### Karnage1993

So you want me to expand (1+x)^3 times (1+x)^3 including the combinations?

If I did that, it would end up being (3C0)*(3C0 + 3C1 + 3C2 + 3C3) etc until I multiplied the other three terms in the first bracket. I still would not know how to simplify even that! What is 3C0 times 3C0 simplified into a combination? The only way I can do that is by getting the actual coefficient, 1, and multiplying it by 1.

Say with 3C2 times 3C2, what would that simplify into using the nCr ?

8. Oct 6, 2011

### Karnage1993

Let's forget part A for a minute. Focusing on part B, what would be my first step into proofing it? Would I expand the 2n Choose n into factorials?

9. Oct 6, 2011

### Dick

What I meant was look at (3C0+3C1*x+3C2*x^2+3C3*x^3)*(3C0+3C1*x+3C2*x^2+3C3*x^3)=(6C0+6C1*x+6C2*x^2+6C3*x^3+6C4*x^4+6C5*x^5+6C6*x^6). The coefficient of x^2 on right side is 6C2=C(3+3,2). To get the coefficient of x^2 of the left side consider all of the products that could give you x^2. That's 3C0*3C2+3C1*3C1+3C2*3C0. Doesn't that look like a). b) follows directly from a) if you put m and l equal to n.