mpdlux

\[ \sum_{k=0}^{0} k^{3}=0^{3}=0 \quad \text{et} \quad \frac{0^{2}(0+1)^{2}}{4}=0 \]

\[ \begin{aligned} \sum_{k=0}^{n+1} k^{3} &= \sum_{k=0}^{n} k^{3} + (n+1)^{3} \\ &= \frac{n^{2}(n+1)^{2}}{4} + (n+1)^{3} \\ &= \frac{n^{2}(n+1)^{2} + 4(n+1)^3}{4} \\ &= \frac{(n+1)^{2}(n^2 + 4n + 4)}{4} \\ &= \frac{(n+1)^{2}(n+2)^{2}}{4} \end{aligned} \]

\[ \forall n \in \mathbb{N}, \quad \sum_{k=0}^{n} k^{3} = \frac{n^{2}(n+1)^{2}}{4} \]

\[ \begin{aligned} \binom{k+1}{\ell+1} – \binom{k}{\ell+1} &= \frac{(k+1)!}{(\ell+1)!(k-\ell)!} – \frac{k!}{(\ell+1)!(k-\ell-1)!} \\ &= \frac{(k+1)k! – (k-\ell)k!}{(\ell+1)!(k-\ell)!} \\ &= \frac{(\ell+1)k!}{(\ell+1)!(k-\ell)!} \\ &= \frac{k!}{\ell!(k-\ell)!} \\ &= \binom{k}{\ell} \end{aligned} \]

\[ \forall k \in \mathbb{N}, \forall \ell \in \llbracket 0, k \rrbracket, \quad \binom{k}{\ell} = \binom{k+1}{\ell+1} – \binom{k}{\ell+1} \]

\[ \sum_{k=\ell}^{n} \binom{k}{\ell} = \sum_{k=\ell}^{n} \left[ \binom{k+1}{\ell+1} – \binom{k}{\ell+1} \right] = \binom{n+1}{\ell+1} – \binom{\ell}{\ell+1} \]

\[ \sum_{k=\ell}^{n} \binom{k}{\ell} = \binom{n+1}{\ell+1} \]

\[ \binom{k}{3} = \frac{k(k-1)(k-2)}{6} \]

\[ \forall k \in \mathbb{N}, \quad \binom{k}{1} = k, \quad \binom{k}{2} = \frac{k(k-1)}{2}, \quad \binom{k}{3} = \frac{k(k-1)(k-2)}{6} \]

\[ \begin{aligned} a\binom{k}{3} + b\binom{k}{2} + c\binom{k}{1} &= a \frac{k(k-1)(k-2)}{6} + b \frac{k(k-1)}{2} + c k \\ &= \frac{a}{6}k^3 + \left( \frac{b}{2} – \frac{a}{2} \right) k^2 + \left( \frac{a}{3} – \frac{b}{2} + c \right)k \end{aligned} \]

\[ \left\{ \begin{array}{l} \frac{a}{6} = 1 \Rightarrow a = 6 \\ \frac{b}{2} – \frac{a}{2} = 0 \Rightarrow b = 6 \\ \frac{a}{3} – \frac{b}{2} + c = 0 \Rightarrow c = 1 \end{array} \right. \]

\[ \forall k \in \mathbb{N}, \quad k^3 = 6\binom{k}{3} + 6\binom{k}{2} + \binom{k}{1} \]

(e) Soit \( n \in \mathbb{N} \). En sommant les égalités obtenues à la question précédente, il vient :

\[ \begin{aligned} \sum_{k=0}^{n} k^{3} &= \sum_{k=0}^{n}\left(6\binom{k}{3}+6\binom{k}{2}+\binom{k}{1}\right)\\ &= 6 \sum_{k=0}^{n}\binom{k}{3} + 6 \sum_{k=0}^{n}\binom{k}{2} + \sum_{k=0}^{n}\binom{k}{1} \\ &= 6 \sum_{k=3}^{n}\binom{k}{3} + 6 \sum_{k=2}^{n}\binom{k}{2} + \sum_{k=1}^{n}\binom{k}{1} \\ &= 6\binom{n+1}{4} + 6\binom{n+1}{3} + \binom{n+1}{2} \end{aligned} \]

D’après la formule du triangle de Pascal généralisée (question 2.(b)). En explicitant les coefficients binomiaux, on obtient :

\[ \begin{aligned} \sum_{k=0}^{n} k^{3} &= 6 \times \frac{(n+1)n(n-1)(n-2)}{24} + 6 \times \frac{(n+1)n(n-1)}{6} + \frac{n(n+1)}{2} \\ &= n(n+1) \cdot \frac{(n-1)(n-2) + 4(n-1) + 2}{4} \\ &= n(n+1) \cdot \frac{n^{2} + n}{4} \\ &= \frac{n^{2}(n+1)^{2}}{4} \end{aligned} \]

Ainsi :

\[ \text{on retrouve bien l’égalité }(*) \]

\[ \begin{aligned} \binom{n}{k}\binom{k}{\ell} &= \frac{n!}{k!(n-k)!} \cdot \frac{k!}{\ell!(k-\ell)!} \\ &= \frac{n!}{(n-k)!\ell!(k-\ell)!} \\ &= \frac{n!}{\ell!(n-\ell)!} \cdot \frac{(n-\ell)!}{(k-\ell)!((n-\ell)-(k-\ell))!} \\ &= \binom{n}{\ell} \binom{n-\ell}{k-\ell} \end{aligned} \]

\[ \forall n, k, \ell \in \mathbb{N}, \quad \ell \leqslant k \leqslant n \Longrightarrow \binom{n}{k}\binom{k}{\ell} = \binom{n}{\ell} \binom{n-\ell}{k-\ell} \]

\[ \begin{aligned} S_{n, \ell} &= \sum_{k=\ell}^{n} (-1)^{k-\ell} \binom{n}{\ell} \binom{n-\ell}{k-\ell} \\ &= \binom{n}{\ell} \sum_{k=\ell}^{n} \binom{n-\ell}{k-\ell} (-1)^{k-\ell} \\ &= \binom{n}{\ell} \sum_{j=0}^{n-\ell} \binom{n-\ell}{j} (-1)^j \quad \text{(changement d’indice \( j = k – \ell \))} \\ &= \binom{n}{\ell} \sum_{j=0}^{n-\ell} \binom{n-\ell}{j} (-1)^j \cdot 1^{n-\ell-j} \\ &= \binom{n}{\ell} \cdot 0^{n-\ell} \end{aligned} \]

\[ S_{n, \ell} = \begin{cases} 0 & \text{si } \ell < n \\ 1 & \text{si } \ell = n \end{cases} \]

\[ \begin{aligned} \sum_{k=0}^{n} (-1)^{n+k} \binom{n}{k} b_k &= \sum_{k=0}^{n} (-1)^{n+k} \binom{n}{k} \sum_{\ell=0}^{k} \binom{k}{\ell} a_\ell \\ &= \sum_{k=0}^{n} \sum_{\ell=0}^{k} (-1)^{n+k} \binom{n}{k} \binom{k}{\ell} a_\ell \\ &= \sum_{0 \leq \ell \leq k \leq n} (-1)^{n+k} \binom{n}{k} \binom{k}{\ell} a_\ell \\ &= \sum_{\ell=0}^{n} (-1)^{n+\ell} a_\ell \sum_{k=\ell}^{n} (-1)^{k-\ell} \binom{n}{k} \binom{k}{\ell} \\ &= \sum_{\ell=0}^{n} (-1)^{n+\ell} a_\ell S_{n, \ell} \end{aligned} \]

\[ \sum_{k=0}^{n} (-1)^{n+k} \binom{n}{k} b_k = \underbrace{\sum_{\ell=0}^{n-1} (-1)^{n+\ell} a_\ell \cdot S_{n, \ell}}_{= 0} + (-1)^{2n} a_n \cdot S_{n, n} = a_n \]

\[ \forall n \in \mathbb{N}, \quad a_n = \sum_{k=0}^{n} (-1)^{n+k} \binom{n}{k} b_k \]

\[ \sum_{k=0}^{0} \binom{0}{k} x_k = \binom{0}{0} x_0 = 1 = 0! \]

\[ \begin{aligned} (n+1)! &= (n+1) \cdot n! \\ &= (n+1) \sum_{k=0}^{n} \binom{n}{k} x_k \\ &= \sum_{k=0}^{n} (n+1) \binom{n}{k} x_k \\ &= \sum_{k=0}^{n} \binom{n+1}{k+1} (k+1) x_k \\ &= \sum_{k=0}^{n} \binom{n+1}{k+1} \left(x_{k+1} – (-1)^{k+1} \right) \\ &= \sum_{\ell=1}^{n+1} \binom{n+1}{\ell} \left( x_\ell – (-1)^\ell \right) \\ &= \sum_{\ell=1}^{n+1} \binom{n+1}{\ell} x_\ell – \sum_{\ell=1}^{n+1} \binom{n+1}{\ell} (-1)^\ell \end{aligned} \]

\[ \sum_{\ell=1}^{n+1} \binom{n+1}{\ell} (-1)^\ell = \sum_{\ell=0}^{n+1} \binom{n+1}{\ell} (-1)^\ell – 1 = 0 – 1 = -1 \]

\[ (n+1)! = \sum_{\ell=1}^{n+1} \binom{n+1}{\ell} x_\ell + 1 = \sum_{\ell=0}^{n+1} \binom{n+1}{\ell} x_\ell \]

\[ \forall n \in \mathbb{N}, \quad n! = \sum_{k=0}^{n} \binom{n}{k} x_k \]

\[ x_n = \sum_{k=0}^{n} (-1)^{n+k} \binom{n}{k} k! = \sum_{k=0}^{n} (-1)^{n+k} \frac{n!}{(n-k)!} = n! \sum_{\ell=0}^{n} \frac{(-1)^{2n – \ell}}{\ell!} \]

\[ \forall n \in \mathbb{N}, \quad x_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} \]