prove the binomial theorem

Answer 1

The Binomial Theorem states that:

(x + y)^n = ∑_(k=0)^n (n choose k) x^(n-k) y^k

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

We can prove this result using mathematical induction.

Base case: n = 0

(x + y)^0 = 1

The formula holds true for n = 0 since (n choose k) = 0 for all k > n or k < 0. Thus the sum collapses to only the term k = 0, which is 1.

Inductive step: Assume that the formula holds true for some positive integer n.

(x + y)^n = ∑_(k=0)^n (n choose k) x^(n-k) y^k

We want to show that the formula holds true for n + 1.

(x + y)^(n+1) = (x + y)(x+y)^n = (x + y)∑_(k=0)^n (n choose k) x^(n-k) y^k

Expanding the product using distributive property:

= ∑_(k=0)^n (n choose k) x^(n-k+1) y^k + ∑_(k=0)^n (n choose k) x^(n-k) y^(k+1)

Notice that in the first sum, we can substitute j = k-1 and obtain:

∑_(j=-1)^(n-1) (n choose j+1) x^(n-j) y^(j+1)

But (n choose j+1) = (n choose j) (n-j)/(j+1) and therefore we can re-write it as:

∑_(j=-1)^(n-1) (n choose j) x^(n-j) y^(j+1) (n-j)/(j+1)

Using the relation that (n choose k) = 0 for k < 0 or k > n, we can simplify the range of summation:

(x + y)^(n+1) = ∑_(k=0)^n (n choose k) x^(n-k+1) y^k + ∑_(k=0)^n (n choose k) x^(n-k) y^(k+1)

= (n choose n) x^(n+1) + ∑_(k=0)^n (n choose k) x^(n-k+1) y^k + ∑_(k=1)^(n+1) (n choose k-1) x^(n-k+1) y^k

= x^(n+1) + ∑_(k=1)^n [(n choose k) + (n choose k-1)] x^(n-k+1) y^k + (n choose n) y^(n+1)

The sum in the middle is the same as:

∑_(k=1)^n (n+1 choose k) x^(n-k+1) y^k

Therefore:

(x + y)^(n+1) = x^(n+1) + ∑_(k=1)^n (n+1 choose k) x^(n-k+1) y^k + y^(n+1)

Which proves the formula for n+1.

Therefore, we have proven the Binomial theorem by mathematical induction.