Lucas Willems

LUCAS WILLEMS

A 21 year-old student passionate about maths and programming

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Newton's binomial formula

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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 !

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