Newton's binomial tells us that :
$$\forall n \in \mathbb{N}, \enspace \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$
$$ (a + b)^n = \sum_{p = 0}^n \binom{n}{p} a^{n-p}b^p \qquad (Newton ) $$
We can use the Pascal's triangle to find binomials coefficients \(\binom{n}{p}\).
$$ (a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b \hspace{0.1em} + \hspace{0.1em} \dots \hspace{0.1em} + \binom{n}{n-2}ab^{n-2} + \binom{n}{n-1}ab^{n-1} +b^n $$
Let show by a recurrence that the statement \((S_n)\) is true:
$$\forall n \in \mathbb{N}, \enspace \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$
$$ (a + b)^n = \sum_{p = 0}^n \binom{n}{p} a^{n-p}b^p \qquad (S_n) $$
Let verify if this is the case for the first term, that is to say when \( n = 0 \).
$$ \sum_{p = 0}^0 \binom{0}{p} a^{0-p}b^p = a^{0 - 0}b^0 = 1 $$
But,
$$ (a + b)^0 = 1 $$
We definitely have:
$$ (a + b)^0 = \sum_{p = 0}^0 \binom{0}{p} a^{0-p}b^p $$
\((S_0)\) is true.
Let \( k \in \mathbb{N} \) be a natural number.
Let us assume that \((S_k)\) is true for all \( k \).
$$ (a + b)^{k} = \sum_{p = 0}^{k} \binom{k}{p} a^{k - p}b^p \qquad (S_{k}) $$
And let verify if it is also the case for \((S_{k + 1})\).
$$ (a + b)^{k + 1} = \sum_{p = 0}^{k + 1} \binom{k + 1}{p} a^{k + 1 - p}b^p \qquad (S_{k + 1}) $$
So that:
$$ (a + b)^{k + 1} = \binom{k + 1}{0}a^{k + 1} + \binom{k + 1}{1}a^{k}b + \binom{k + 1}{2}a^{k-1}b^2 \ + \ ... \ + \ \binom{k + 1}{k}ab^{k} + \binom{k + 1}{k + 1}b^{k + 1} \qquad (S_{k + 1}') $$
Multiplying \(( S_{k} ) \) by \(( a + b) \), we do have:
$$ (a + b)^{k}(a + b) = (a + b) \times \sum_{p = 0}^k \binom{k}{p} a^{k-p}b^k $$
$$ (a + b)^{k + 1} = (a + b) \times \left[ \binom{k}{0}a^k + \binom{k}{1}a^{k - 1}b + \binom{k}{2}a^{k - 2}b^2 \ + \ ... \ + \ \binom{k}{k - 1}ab^{k -1} + \binom{k}{k} b^k \right] $$
$$ (a + b)^{k + 1} = \binom{k}{0} \left(a^{k + 1} + ba^k \right) + \binom{k}{1}\left(a^kb + a^{k - 1}b^2 \right) \ + \ ... \ + \ \binom{k}{k - 1}\left(a^2b^{k -1} + ab^k \right) + \binom{k}{k}\left(ab^k + b^{k + 1}\right) $$
$$ (a + b)^{k + 1} = \textcolor{#606B9E}{\binom{k}{0}} a^{k + 1} + \textcolor{#446e4f}{\left[\binom{k}{0} + \binom{k}{1} \right]} ba^k + \textcolor{#A65757}{\left[\binom{k}{1} + \binom{k}{2} \right]} a^{k - 1}b^2 \ + \ ... \ + \ \textcolor{#7C578A}{\left[\binom{k}{k -1} + \binom{k}{k} \right]}ab^k + \textcolor{#606B9E}{\binom{k}{k}} b^{k +1} $$
But we know thanks to the Pascal's formula that :
$$ \forall (p,n) \in \hspace{0.05em}\mathbb{N}^2, \enspace p \leqslant n -1, $$
$$ \binom{n}{p} = \binom{n - 1}{p - 1} + \binom{n - 1}{p} \qquad (Pascal) $$
And as well:
$$ \binom{n + 1}{p + 1} = \binom{n}{p} + \binom{n}{p + 1} \qquad (Pascal^*) $$
Consequently, thanks to \( (Pascal^*) \), we do have:
$$ (a + b)^{k + 1} = \textcolor{#606B9E}{\binom{k}{0}} a^{k + 1} + \textcolor{#446e4f}{\binom{k + 1}{1} } ba^k + \textcolor{#A65757}{\binom{k + 1}{2}} a^{k - 1}b^2 \ + \ ... \ + \ \textcolor{#7C578A}{\binom{k + 1}{k}}ab^k + \textcolor{#606B9E}{\binom{k}{k }} b^{k +1} $$
Now, we notice that:
$$ \binom{k}{0} = \binom{k + 1}{0} $$
And:
$$ \binom{k}{k} = \binom{k + 1}{k + 1} $$
And finally we do have:
$$ (a + b)^{k + 1} = \binom{k + 1}{0} a^{k + 1} + \binom{k + 1}{1} ba^k + \binom{k + 1}{2} a^{k - 1}b^2 \ + \ ... \ + \ \binom{k + 1}{k}ab^k + \binom{k + 1}{k + 1 } b^{k +1} $$
If we write it under another form, we do have the beginning statement \(( S_{k + 1} ) \):
$$ (a + b)^{k + 1} = \sum_{p = 0}^{k + 1} \binom{k + 1}{p} a^{k + 1 - p}b^p \qquad (S_{k + 1}) $$
Thus, \((S_{k + 1})\) is true.
The statement \((S_n)\) is true for its first terme \(n_0 = 0\) and it is hereditary from terms to terms for all \(k \in \mathbb{N}\).
By the recurrence principle, that statement is true for all \(n \in \mathbb{N}\).
Nous venons de prouver par une récurrence que :
$$\forall n \in \mathbb{N}, \enspace \forall (a, b) \in \hspace{0.05em} \mathbb{R}^2,$$
$$ (a + b)^n = \sum_{p = 0}^n \binom{n}{p} a^{n-p}b^p \qquad (Newton ) $$
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