Summation of Fibonacci numbers $F_n$ with $n$ odd vs. even Compare the summation below:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i-1}}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\
&=1+2+5+\cdots+F_{2n-1}\\
&=F_{2n}\\
\end{align}
$$
with this one:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i}}&=F_2+F_4+F_6+\cdots+F_{2n}\\
&=1+3+8+\cdots+F_{2n}\\
&=F_{2n+1}-1\\
\end{align}$$
When I first discovered these patterns I was amazed. Naively I had thought that an every-other-number sum of Fibonacci numbers would be the same pattern whether the parity of their indices was odd or even, but I was wrong! Why is the above true, where the summation of odd-indexed Fibonacci numbers is another Fibonacci number, but the even-indexed sum is a Fibonacci number minus 1?

Here is a late answer using my favorite Fibonacci technique.
The matrix formulation for the Fibonacci sequence is well worth knowing and easily proved:
$$
\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=
\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}
$$
Let
$
A=\begin{pmatrix}1&1\\1&0\end{pmatrix}
$.
Since the second column of $A^{2k}$ is $\begin{pmatrix}F_{2k}\\F_{2k-1}\end{pmatrix}$,
the second column of
$$
S = A^2 + A^4 + \cdots + A^{2n}
$$ 
contains exactly the sums we're interested in.
We have $(A^2-I)S=A^{2n+2}-A^2$.
Since $A^2=A+I$ (the Fibonacci recurrence!), we get $AS=A^{2n+2}-A^2$ and so
$$
S=A^{2n+1}-A
=\begin{pmatrix}F_{2n+2}&F_{2n+1}\\F_{2n+1}&F_{2n}\end{pmatrix}
-\begin{pmatrix}1&1\\1&0\end{pmatrix}
=\begin{pmatrix}*&F_{2n+1}-1\\*&F_{2n}\end{pmatrix}
$$ 
as claimed. This also explains where the $-1$ comes from.
