Golden Ratio.. or is it?

For the first three Fibonnaci numbers, i.e. $f_{0},f_{1},f_{2}$ the numbers are $even, odd, odd$ Then we try to find the next three numbers: $odd+odd=even$ $odd+even=odd$ $even+odd=odd$ This will continue recursively and thus, we conclude that $f_{n}$ is even if and only if $n$ is divisible by 3. And so, $4+9+6+2=21$ Hence, $f_{4962}$ is even.

In fact, since 4962 is divisible by 3, f(4962) is divisible by f(3), and f(3) is 2. (Property of fibonnaci series)