We fix $n \in \mathbb { N }$. We define the linear map: $$\begin{aligned}
\Delta : \mathbb { R } [ X ] & \rightarrow \mathbb { R } [ X ] \\
P ( X ) & \mapsto P ( X + 1 ) - P ( X )
\end{aligned}$$ We denote $F = \left\{ P \in \mathbb { R } _ { n } [ X ] \mid P ( 0 ) = 0 \right\}$, then $G = \operatorname { Vect } \left( X ^ { 2 k + 1 } ; 0 \leqslant k \leqslant n - 1 \right)$. Let $Q ( X )$ be the polynomial such that $\forall p \in \mathbb { N } , Q ( p ) = \sum _ { k = 0 } ^ { p } k$. V.D.1) Recall the explicit expression of the polynomial $Q ( X )$. V.D.2) Show that the map: $$\begin{aligned}
\Phi : F & \rightarrow G \\
P ( X ) & \mapsto \Delta ( P ( Q ( X - 1 ) ) )
\end{aligned}$$ is an isomorphism. V.D.3) Deduce that for all $r \in \mathbb { N }$, there exists a unique polynomial $P _ { r } ( X )$ such that $$\forall p \in \mathbb { N } , \quad \sum _ { k = 1 } ^ { p } k ^ { 2 r + 1 } = P _ { r } \left( \frac { p ( p + 1 ) } { 2 } \right)$$
We fix $n \in \mathbb { N }$. We define the linear map:
$$\begin{aligned}
\Delta : \mathbb { R } [ X ] & \rightarrow \mathbb { R } [ X ] \\
P ( X ) & \mapsto P ( X + 1 ) - P ( X )
\end{aligned}$$
We denote $F = \left\{ P \in \mathbb { R } _ { n } [ X ] \mid P ( 0 ) = 0 \right\}$, then $G = \operatorname { Vect } \left( X ^ { 2 k + 1 } ; 0 \leqslant k \leqslant n - 1 \right)$.
Let $Q ( X )$ be the polynomial such that $\forall p \in \mathbb { N } , Q ( p ) = \sum _ { k = 0 } ^ { p } k$.
V.D.1) Recall the explicit expression of the polynomial $Q ( X )$.
V.D.2) Show that the map:
$$\begin{aligned}
\Phi : F & \rightarrow G \\
P ( X ) & \mapsto \Delta ( P ( Q ( X - 1 ) ) )
\end{aligned}$$
is an isomorphism.
V.D.3) Deduce that for all $r \in \mathbb { N }$, there exists a unique polynomial $P _ { r } ( X )$ such that
$$\forall p \in \mathbb { N } , \quad \sum _ { k = 1 } ^ { p } k ^ { 2 r + 1 } = P _ { r } \left( \frac { p ( p + 1 ) } { 2 } \right)$$