BinomialCoefficient

The Binomial Theorem

Theorem (Binomial Theorem): Let $n$ be a non-negative integer and $a, b$ real or complex numbers. Then, the power of the binomial $(a + b)^n$ can be expanded in the form of a sum as follows:

$$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k $$

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ denotes the binomial coefficient.

Proof

We prove the theorem by induction on $n$.

Base case: For $n=0$, $(a + b)^0 = 1$. The right-hand side is $\sum_{k=0}^{0} \binom{0}{k} a^{0-k} b^k = \binom{0}{0} a^0 b^0 = 1$, which confirms the base case.

Inductive step: Assume the statement holds for some non-negative integer $n$, i.e.,

$$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. $$

We need to show that the statement holds for $n+1$, that is,

$$ (a + b)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k} b^k. $$

Starting from the left-hand side, we have:

$$ (a + b)^{n+1} = (a + b)(a + b)^n = (a + b) \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k. $$

Expanding the product, we get:

$$ \sum_{k=0}^{n} \binom{n}{k} a^{n+1-k} b^k + \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k+1}. $$

Rearranging terms and applying the identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$, we arrive at:

$$ \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k} b^k, $$

which completes the inductive step and thus proves the theorem.