Prove the binomial theorem.

The binomial theorem states that (a+b)^n = Σ(n choose k) * a^(n-k) * b^k.

The binomial theorem is a formula that allows us to expand expressions of the form (a+b)^n, where n is a positive integer. The theorem states that:

(a+b)^n = Σ(n choose k) * a^(n-k) * b^k

where Σ denotes the sum over all values of k from 0 to n, and (n choose k) is the binomial coefficient, given by:

(n choose k) = n! / (k! * (n-k)!)

To prove the binomial theorem, we can use mathematical induction. The base case is when n=1, in which case we have:

(a+b)^1 = a+b

which is clearly true. Now suppose that the theorem holds for some positive integer n, so that:

(a+b)^n = Σ(n choose k) * a^(n-k) * b^k

We want to show that the theorem also holds for n+1, i.e. that:

(a+b)^(n+1) = Σ((n+1) choose k) * a^((n+1)-k) * b^k

To do this, we can use the binomial theorem again, but with (a+b)^(n+1) expanded as (a+b)*(a+b)^n:

(a+b)^(n+1) = (a+b)*(a+b)^n
= a*(a+b)^n + b*(a+b)^n

Now we can use the induction hypothesis to expand each of these terms:

a*(a+b)^n = a*Σ(n choose k) * a^(n-k) * b^k
= Σ(n choose k) * a^(n-k+1) * b^k

b*(a+b)^n = b*Σ(n choose k) * a^(n-k) * b^k
= Σ(n choose k) * a^(n-k) * b^(k+1)

Adding these two expressions together, we get:

(a+b)^(n+1) = Σ(n choose k) * a^(n-k+1) * b^k + Σ(n choose k) * a^(n-k) * b^(k+1)

We can