Fibonacci Number by Power of 2/Proof 2

Proof
Setting $j = 2 k + 1$ for $0 \le k \le \left({j - 1}\right) / 2$ gives:


 * $\displaystyle \sum_{k \mathop \ge 0} \binom n {2 k + 1} 5^k$

and the limits of the index of the summation are irrelevant, as $\dbinom n {2 k + 1} = 0$ for $j < 0$ and $j > n$.

Hence the result.