If $f (x +1) + f (x −1) = 2 f (x)$ and $f(0) = 0$, then $f(n), n ∈ N$, is

$nf(1)$

$f (x +1) + f (x −1) = 2 f (x) , f (0) = 0$

Putting x = 1, $f(2) + f(0) = 2f(1), ∴ f(2) = 2f(1)$

Let $f(k) = kf(1)$ for $k ≤ n$

Then $f(n + 1) + f(n − 1) = 2f(n)$

$⇒ f (n +1) = 2(nf (1)) − f (n −1) = 2nf (1) − (n −1) f (1) = (n +1) f (1)$

$∴ f(n) = nf(1)$ for all $n ∈ N.$