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Binomial theorem

The binomial theorem states that $\forall x,y \in \mathbb{R}$ and $\forall n, k \in \mathbb{N}$, such as $k \leq n$, we have:

[math] \left ( x+y \right )^n = \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k [/math]

Inductive Proof[edit]

Prerequisite information[edit]

The proof[edit]

In order to show that the binomial theorem is true for $n \in \mathbb{N}$. We will first prove it is true for $n=0$, then for $n \in \mathbb{N}^*$ by induction.

Proof for $n = 0$[edit]

We have:

[math](x+y)^0 = 1[/math]

and

[math] \sum_{k=0}^{0}\binom{n}{k}x^{n-k}y^{k}=\binom{0}{0}x^0y^0=1[/math]

We then deduce that the binomial theorem is true for $n=0$.

Inductive step[edit]

We know that for [math]n =0[/math], we have:

[math] \left ( x+y \right )^n = \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k [/math]

For $n+1$, we have:

[math] \begin{align*} \left(x+y\right)^{n+1} & = \left(x+y\right)\left(x+y\right)^{n}\\ &= x\left(x+y\right)^{n}+y\left(x+y\right)^{n}\\ &= x\left ( \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k \right ) + y\left ( \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^k \right ) \\ &= \sum_{k=0}^{n}\binom{n}{k}x^{n-k+1}y^k + \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k+1} \\ &= \sum_{k=0}^{n}\binom{n}{k}x^{n-k+1}y^k + \sum_{k=1}^{n+1}\binom{n}{k-1}x^{n-k+1}y^{k} \\ &= \binom{n}{0}x^{n+1}y^0 + \left[ \sum_{k=1}^{n}\binom{n}{k}x^{n-k+1}y^k + \sum_{k=1}^{n}\binom{n}{k-1}x^{n-k+1}y^{k} \right ]+ \binom{n}{n}x^{0}y^{n+1} \\ &= x^{n+1} + \left[\sum_{k=1}^{n}\binom{n}{k}x^{n-k+1}y^k + \sum_{k=1}^{n}\binom{n}{k-1}x^{n-k+1}y^{k} \right]+ y^{n+1} \\ &= x^{n+1} + \left[ \sum_{k=1}^{n} \left (\binom{n}{k} + \binom{n}{k-1} \right)x^{n-k+1}y^{k} \right] + y^{n+1} \\ &= x^{n+1} + \left[ \sum_{k=1}^{n} \binom{n+1}{k} x^{n-k+1}y^{k} \right ]+ y^{n+1} \\ \end{align*} [/math]

This is why we have:

[math] \begin{align*} \left(x+y\right)^{n+1} &= \sum_{k=0}^{n+1} \binom{n+1}{k} x^{n-k+1}y^{k} \end{align*} [/math]

Because we know that the binomial theorem is true for $n=0$ and for $n=n+1$, we deduce by induction that it is true for $n\in\mathbb{N}$.