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. :)

The following is formally not correct because it uses the "$\cdots$" symbol but it gives some insight. The proof can be formalized using induction. This was already done by @Jytug.
$$\begin{eqnarray}
F(1) + F(3) + F(5) + \cdots + F(2n-1)&= \\
0+\left(F(1) + F(3) + F(5) + \cdots  + F(2n-1)\right) &= \\
F(0)+\left(F(1) + F(3) + F(5) + \cdots  + F(2n-1)\right)&=\\
\left(F(0)+F(1)\right) +\left( F(3) + F(5) + \cdots + F(2n-1)\right)&=\\
F(2)+\left( F(3) + F(5) + \cdots  + F(2n-1)\right)&=\\
\left(F(2)+F(3)\right) + \left( F(5) + \cdots  + F(2n-1)\right)&=\\
F(4) + \left( F(5) + \cdots  + F(2n-1)\right)&=\\
\cdots&=\\
F(2n-2)+F(2n-1)&=\\
F(2n)&
\end{eqnarray}
$$
