Prove that $F(1) + F(3) + F(5) + ... + F(2n-1) = F(2n)$ (These are Fibonacci numbers; $f(1) = 0$, $f(3) = 1$, $f(5) = 5$, etc.) I'm having trouble proving this with induction, I know how to prove the base case and present the induction hypothesis but I'm unfamiliar with proving series such as this. Any help would be great. :)

Here is a solution using Binet's formula:
$F(n) = a \alpha^n + b \beta^n$, for some $a,b \in \mathbb R$, and $\alpha,\beta$ the roots of $X^2=X+1$. (Their precise value is not really important here.)
Now,
$
\alpha^1 + \alpha^3 + \cdots + \alpha^{2n-1}
=\alpha(1+\alpha^2 + \alpha^4 + \cdots + \alpha^{2n-2})
=\alpha\dfrac{\alpha^{2n}-1}{\alpha^2-1}
=\alpha^{2n}-1
$
because $\alpha=\alpha^2-1$.
Analogously,
$
\beta^1 + \beta ^3 + \cdots + \beta ^{2n-1}
=\beta ^{2n}-1
$.
Therefore,
$
F(1) + F(3) + F(5) + \cdots + F(2n-1) = a(\alpha^{2n}-1)+b(\beta ^{2n}-1)=F(2n)-a-b
$.
Finally, $a+b=F(0)=0$.
