# LUCAS WILLEMS

A 23 year-old student passionate about maths and programming

# Newton's binomial formula

## Article

Newton's binomial formula is the following :

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

where $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ can be seen as combinatorics or as coefficients in Pascal's triangle.

Note : you need to have $$ab = ba$$ that is the case if $$(a, b) \in \mathbb{C}^2$$.

This formula generalize the calculation of $$(a+b)^2$$, $$(a+b)^3$$... to the power $$n$$ and is easily conjecturable by calculating for $$n = 2$$, then $$n = 3$$... but much more difficult to prove. So, the purpose of this article is to show you how to prove this formula.

Summary

## Proof

To prove this formula, let's use induction with this statement :

$$\forall n \in \mathbb{N} \qquad H_n : (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

that leads us to the following reasoning :

Bases :
For $$n = 0$$, $$(a+b)^0 = 1 = \binom{0}{0} a^0 b^0$$. So, $$H_0$$ holds.

Induction steps :
For $$n+1$$ :

$$(a+b)^{n+1} = (a+b)(a+b)^n$$

As we assume $$H_n$$ holds, we have :

\begin{align} (a+b)^{n+1} &= (a+b) \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k = a \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k + b \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \\ &= \sum_{k=0}^{n} \binom{n}{k} a^{n-k+1} b^k + \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k+1} \\ &= \sum_{k=0}^{n} \binom{n}{k} a^{n-k+1} b^k + \sum_{k=1}^{n+1} \binom{n}{k-1} a^{n-(k-1)} b^{(k-1)+1} \\ &= \binom{0}{0}a^{n+1}b^0 + \sum_{k=1}^{n} \binom{n}{k} a^{n+1-k} b^k + \binom{n}{n}a^0b^{n+1} + \sum_{k=1}^{n} \binom{n}{k-1} a^{n+1-k} b^{k} \\ &= \binom{0}{0}a^{n+1}b^0 + \binom{n+1}{n+1}a^0b^{n+1} + \sum_{k=1}^{n} \left(\binom{n}{k} + \binom{n}{k-1}\right)a^{n+1-k} b^k \end{align}

As Pascal's triangle allows us to see easily that $$\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$$, we get :

\begin{align} (a+b)^{n+1} &= \binom{0}{0}a^{n+1}b^0 + \binom{n+1}{n+1}a^0b^{n+1} + \sum_{k=1}^{n}\binom{n+1}{k}a^{n+1-k} b^k \\ &= \sum_{k=0}^{n+1}\binom{n+1}{k}a^{n+1-k}b^k\end{align}

Furthermore, $$H_n \Rightarrow H_{n+1}$$.

Conclusion :
$$H_n$$ holds $$\forall n \in \mathbb{N}$$.

We have just proved Newton's binomial formula !