<s>[INST] 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?
 [/INST] 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.
</s>