Let $n\in {1,...,N}$ and
$q(n)=\frac{1}{2}(q(n-1)+q(n+1))$.
Furthemore $q(0)=1$ and $q(N+1)=1$. Prove that $q(n)=1$ for all $n\in \{1,...,N\}$. My first idea was to try induction over N and I can do the base case N=1. But the induction step is difficult. Is there an easier way to prove this or do I have to do it by induction?
To give a complete answer using the hints from Anurag A, consider the sequence $s(n) = q(n) - q(n-1)$. We can rearrange the given relation to be: $$ 2q(n) - q(n-1) = q(n+1) \quad \implies \quad q(n) - q(n-1) = q(n+1) - q(n) \quad \implies \quad s(n) = s(n+1) $$ This holds for all $n \in \{1, 2, \ldots, N\}$, so $s: \{1, 2, \ldots N\} \to \mathbb{R}$ is a constant sequence, say $s(n) = C$ for all $n \in \{1, 2, \ldots N\}$. Then, by considering the sum of the telescoping series, we have: $$ C \cdot N = \sum_{n = 1}^N s(n) = \sum_{n = 1}^N q(n) - q(n-1) = -q(0) + q(N) = -1 + 1 = 0 $$ Thus, we have that $C \cdot N = 0$, and by the structure of your question we assume that $N \neq 0$ which means we must have $C = 0$. Therefore, $s$ is the identically zero sequence which means that $q(n) = q(n-1)$ for all $n \in \{1, 2, \ldots, N\}$. $\square$