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?
 A: Fibonacci numbers are defined by the recurrence relation,
$$F_n=F_{n-1}+F_{n-2},~~~F_1=F_2=1.$$
Rearranging, we have $F_{n-1}=F_n-F_{n-2}$. Letting $n=2k$,
$$F_{2k-1}=F_{2k}-F_{2(k-1)},$$
hence, the sum of odd-indexed Fibonacci numbers telescopes:
$$\sum_{k=2}^{m}F_{2k-1}=\sum_{k=2}^{m}(F_{2k}-F_{2(k-1)})=F_{2m}-F_{2}.$$
Since $F_1=F_2$,
$$\sum_{k=2}^{m}F_{2k-1}=F_{2m}-F_{2}\\
\implies F_1+\sum_{k=2}^{m}F_{2k-1}=F_{2m}\\
\implies \sum_{k=1}^{m}F_{2k-1}=F_{2m},$$
which is the formula the OP mentioned finding.
The derivation of the analogous formula for a sum of even-indexed Fibonacci numbers is highly similar. The key is the recurrence relation.
A: A clean way to see this is by using generating functions. Define $F(z) = \sum_{n \ge 0} F_n z^n$, take the recurrence:
$$
F_{n + 2} = F_{n + 1} + F_n \qquad F_0 = 0, F_1 = 1
$$
Multiply by $z^n$, sum over all valid values for $n$, i.e., $n  \ge 0$, and recognize the resulting sums:
$$
\frac{F(z) - F_0 - F_1 z}{z^2}
  = \frac{F(z) - F_0}{z} + F(z)
$$
Solving:
$$
F(z) = \frac{z}{1 - z - z^2}
$$
We also have, if $A(z) = \sum_{n \ge 0} a_n z^n$ then:
\begin{align}
\sum_{n \ge 0} a_{2 n} z^{2 n}
  &= \frac{A(z) + A(-z)}{2} \\
\sum_{n \ge 0} a_{2 n + 1} z^{2 n + 1}
  &= \frac{A(z) - A(-z)}{2} \\
\sum_{n \ge 0} \left( \sum_{0 \le k \le n} a_k \right) z^n
  &= \frac{A(z)}{1 - z}
\end{align}
So, for even/odd Fibonacci numbers:
\begin{align}
F_e(z)
  &= \sum_{n \ge 0} F_{2 n} z^n \\
  &= \frac{F(z^{1/2}) + F(- z^{1/2})}{2} \\
  &= \frac{z}{1 - 3 z + z^2} \\
F_o(z)
  &= \sum_{n \ge 0} F_{2 n + 1} z^n \\
  &= \frac{F(z^{1/2}) - F(- z^{1/2})}{2 z^{1/2}} \\
  &= \frac{1 - z}{1 - 3 z + z^2} \\
\end{align}
\begin{align}
\sum_{n \ge 0} \left( \sum_{0 \le k \le n} F_{2 n} \right) z^n
  &= \frac{F(z^{1/2}) + F(- z^{1/2})}{2 (1 - z)} \\
  &= \frac{z}{(1 - z) (1 - 3z + z^2)} \\
  &= \frac{1 - z}{1 - 3 z + z^2} - \frac{1}{1 - z}
\end{align}
The first term is the generating function of the odd Fibonacci numbers, the second one is the generating function of the sequence of ones. Comparing coefficients:
$$
\sum_{0  \le k \le n} F_{2 n} = F_{2 n + 1} - 1
$$
Similarly, as $F_0 = 0$:
\begin{align}
\sum_{n \ge 0} \left( \sum_{0 \le k \le n} F_{2 n + 1} \right) z^n
  &= \frac{F(z^{1/2}) - F(- z^{1/2})}{2 z^{1/2} (1 - z)} \\
  &= \frac{1}{1 - 3z + z^2} \\
  &= \frac{F_e(z) - F_0}{z}
\end{align}
The last expression corresponds to the even Fibonacci numbers shifted by one:
$$
\sum_{0 \le k \le n} F_{2 n + 1} = F_{2 n + 2}
$$
Note that we didn't need any premonition on what the sums would turn out to be.
A: Here's a slightly sneaky way to remove the disparity between the two sums:  Use the indexing convention $F_0=F_1=1$.  You'll then find that
$$\sum_{i=1}^n F_{2i-1}=F_{2n}-1$$
and
$$\sum_{i=1}^n F_{2i}=F_{2n+1}-1$$
Voila! The pattern is the same for both!
Added later:  Here's another way to make the two patterns look alike, without playing around with the indexing convention (indeed, it doesn't matter what convention you use):
$$F_1+F_3+\cdots+F_{2n-1}=F_{2n}-F_0$$
and
$$F_2+F_4+\cdots+F_{2n}=F_{2n+1}-F_1$$
In general, if $m$ and $n$ have the same parity and $m\lt n$, then
$$F_m+F_{m+2}+\cdots+F_n=F_{n+1}-F_{m-1}$$
A: 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.
A: By adding the two summations together, you get:
$$\begin{align}
\sum_{i=1}^n F_{2i-1}+\sum_{i=1}^n F_{2i}&=F_{2n}+F_{2n+1}-1\\
&=F_{2n+2}-1\\
&=\sum_{i=1}^{2n}F_i\\
\end{align}$$
In this way, both patterns can be mathematically justified—although the mechanics behind this phenomenon are still somewhat of a mystery to me.
Update: The two formulas can be merged into one in the following way:
$$\begin{align}
0&=F_0\\
1&=F_1\\
\smash[b]{\sum_{i=1}^n F_{2i-1}}&=F_{2n}\\
&=F_{2n+0}-0\\
&=F_{2n+0}-F_0\\
\smash[b]{\sum_{i=1}^n F_{2i}}&=F_{2n+1}-1\\
&=F_{2n+1}-F_1\\
\therefore\smash[b]{\sum_{i=1}^n F_{2i+r-1}}&=F_{2n+r}-F_r &r=\begin{cases}
0, &\text{if $2i+r-1$ is odd}\\
1, &\text{if $2i+r-1$ is even}
\end{cases}
\end{align}$$
