# Timeless Theorems of Mathematics/Binomial Theorem

The Binomial Theorem is a fundamental theorem in algebra that provides a
formula to expand powers of binomials. It allows us to easily expand
expressions, $(a+b)^n$, where $a$ and $b$ are real numbers or variables,
and $n\geq0$.

## Proof

**Proposition:** For any real numbers $a, b$ and $n\ge0,$
$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k$ Where
$\binom{n}{k} = \frac{n!}{k! (n-k)!}$

**Proof *(Mathematical Induction)*:**

For $n = 0, (a + b)^n = (a + b)^0 = 1$

For $n = 1, (a + b)^n = (a + b)^1 = (a + b)$

For $n = 2, (a + b)^n = (a + b)^2$ $= a^2 + 2ab + b^2$
$= \binom{2}{0}a^2b^0 + \binom{2}{1}a^1b^1 + \binom{2}{2}a^0b^2$
$=\sum_{k=0}^2 \binom{2}{k}a^{2-k}b^k$

For $n = 3, (a + b)^n = (a + b)^3$ $= a^3 + 3a^2b + 3ab^2 + b^3$
$= \binom{3}{0}a^3b^0 + \binom{3}{1}a^2b^1 + \binom{3}{2}a^1b^2 + \binom{3}{3} a^0b^3$
$=\sum_{k=0}^3 \binom{3}{k}a^{3-k}b^k$

Let\'s assume $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k$ for
some $a\ge0.$ Now we just have to show that the equation holds true for
$n + 1.$

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

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

$= (\binom {n}{0} a^nb^0 + \binom {n}{1} a^{n-1}b^1 + \binom {n}{2} a^{n-2}b^2 + ... + \binom {n}{n} a^0b^n) (a + b)$

$= a(\binom {n}{0} a^nb^0 + \binom {n}{1} a^{n-1}b^1 + \binom {n}{2} a^{n-2}b^2 + ... + \binom {n}{n} a^0b^n) + b(\binom {n}{0} a^nb^0 + \binom {n}{1} a^{n-1}b^1 + \binom {n}{2} a^{n-2}b^2 + ... + \binom {n}{n} a^0b^n)$

$= (\binom {n}{0} a^{n+1}b^0 + \binom {n}{1} a^{n}b^1 + \binom {n}{2} a^{n-1}b^2 + ... + \binom {n}{n} ab^n) + (\binom {n}{0} a^nb^1 + \binom {n}{1} a^{n-1}b^2 + \binom {n}{2} a^{n-2}b^3 + ... + \binom {n}{n} a^0b^{n+1})$

$= \binom {n}{0} a^{n+1}b^0 + a^{n}b^1 (\binom {n}{1} + \binom {n}{0}) + a^{n-1}b^2 (\binom {n}{2} + \binom {n}{1}) + ... + ab^n (\binom {n}{n} + \binom {n}{n-1}) + \binom {n}{n} a^0b^{n+1})$

$= \binom {n+1}{0} a^{n+1}b^0 + \binom {n+1}{1}a^{n}b^1 + \binom {n+1}{2} a^{n-1}b^2 + ... + \binom {n + 1}{n} ab^n + \binom {n+1}{n+1} a^0b^{n+1})$

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