Problem 4. Let $\left(a_{n}\right)_{n \in N}$ be a sequence of real numbers with the property that for any $m, n \in N, m \geq n$,

$2\left(a_{m+n}+a_{m-n}\right)=a_{2 m}+a_{2 n}$ and $a_{1}=1$. Determine a formula for the general term of the sequence $\left(a_{n}\right)_{n \in N}$.
Solution:

$$
\begin{aligned}
& m=n=0 \Rightarrow a_{0}=0 \\
& m=1, n=0 \Rightarrow a_{2}=4 \\
& \mathbf{1 p} \\
& m=n+2 \Rightarrow 2\left(a_{2 n+2}+a_{2}\right)=a_{2 n+4}+a_{2 n} \Rightarrow \\
& \Rightarrow 2 a_{2 n+2}+8=a_{2 n+4}+a_{2 n} \\
& \mathbf{1 p} \\
& n=0 \Rightarrow 4 a_{m}=a_{2 m} \\
& 8 a_{n+1}+8=4 a_{n+2}+4 a_{n} \Rightarrow \\
& a_{n+2}-a_{n+1}=a_{n+1}-a_{n}+2 \\
& \mathbf{2 p}
\end{aligned}
$$

Let $b_{n}=a_{n+1}-a_{n} \Rightarrow (b_{n})$ is an arithmetic progression, $\quad b_{1}=3, r=2 \Rightarrow b_{n}=2 n+1 \Rightarrow$ 1 p

$$
\begin{aligned}
& a_{n+1}-a_{n}=2 n+1 \Rightarrow \sum_{k=0}^{n-1}\left(a_{k+1}-a_{k}\right)=\sum_{k=0}^{n-1}(2 k+1) \Rightarrow \\
& a_{n}-a_{0}=n(n-1)+n \Rightarrow a_{n}=n^{2}, \forall n \in N . \mathbf{( 2 p )}
\end{aligned}
$$