11.2. Let the sequence $\left(a_{n}\right)_{n=0}^{\infty}$ be such that $a_{1}=1$ and $a_{m+n}+a_{m-n}=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right), \quad \forall m \geq n \geq 0$. Determine $a_{2020}$.
## Solution.

Considering $m=n$, we have $a_{2 m}+a_{0}=\frac{1}{2}\left(a_{2 m}+a_{2 m}\right)=a_{2 m}, \quad \forall m \geq 0$, which implies $a_{0}=0$.

Similarly, considering $n=0$, we get $a_{m}+a_{m}=\frac{1}{2}\left(a_{2 m}+a_{0}\right)=\frac{1}{2} a_{2 m}$, i.e., $a_{2 m}=4 a_{m}, \forall m \geq 0$.

Next, for $m=n+2, \forall n \geq 0$, we obtain

$$
4 a_{n+1}+4=4 a_{n+1}+4 a_{1}=a_{2 n+2}+a_{2}=a_{m+n}+a_{m-n}=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right)=\frac{1}{2}\left(4 a_{m}+4 a_{n}\right)=2\left(a_{n+2}+a_{n}\right)
$$

Thus, $a_{n+2}=2 a_{n+1}-a_{n}+2, \forall n \geq 0$, where $a_{0}=0$ and $a_{1}=1$.

We prove by mathematical induction the formula $a_{n}=n^{2}, \forall n \geq 0$. For $n<2$ the equality holds, since $a_{0}=0=0^{2}$ and $a_{1}=1=1^{2}$. Suppose the formula holds for $\forall n \leq N$. We have

$$
a_{N+1}=2 a_{N}-a_{N-1}+2=2 N^{2}-(N-1)^{2}+2=(N+1)^{2}
$$

Thus, the statement is also true for $n=N+1$. By the method of mathematical induction, we have $a_{n}=n^{2}, \forall n \geq 0$. We obtain $a_{2020}=2020^{2}$.