discrete math - binomial coefficients

Question

Find three integers a, b, and c such that aC(m,3) + bC(m,2) + cC(m,1) = m³. Use this fact to compute the sum 1³ + 2³ +...+ n³.

I have gotten the first part, a =6, b = 6, c = 1 but I am stuck on the second part. Thank you!

Answer 1

We want to compute the sum ∑m³ from m=1 to m=n.

∑m³ = a∑C(m,3) + b∑C(m,2) + c∑C(m,1)

Also, I'm guessing that you already know the identity for summing over the top coordinate of the binomial. It is the identity (sometimes called the "hockey stick identity")

∑C(m,r) = C(n+1,r+1) with the sum taken from m=0 to m=n.

We can almost use this formula on, for instance, ∑C(m,3), the only problem is that the index here starts at m=1 and not m=0. But actually that turns out not to be such a problem anyway since C(0,3) is just the number 0. So it's actually not even really a term. Therefore

∑m³ = a∑C(m,3) + b∑C(m,2) + c∑C(m,1)

= a * C(n+1,4) + b * C(n+1,3) + c * C(n+1,2).

You can also test out whether this answer is correct. I entered the following into WolframAlpha:

6*(11 choose 4) + 6*(11 choose 3) + 11 choose 2 = sum m^3 from 1 to 10

It checks the result for when n=10.