<s>[INST] Prove that the sum of the first $n$ even terms of the Fibonacci sequence is given by $(F_{3n} + F_{3n + 3} - 2) / 4$ not using induction? each even term in the Fibonacci sequence has a position index which is a multiple of 3, therefore the even term is $F_{3n}$:
$F_{3 (0)} = 0$
$F_{3 (1)} = 2$
$F_{3 (2)} = 8$
$F_{3 (3)} = 34$
$F_{3 (4)} = 144$
...
How can I prove and reconstruct the formula which states that the sum of the first $n$ even terms (2, 8, 34, ...) of the Fibonacci sequence is given by $(F_{3n} + F_{3n + 3} - 2)/4$ not using induction?
For example, given:
$U_{n} = F_{3n}$
$S_{n} = U_{0} + U_{1} + U_{2} + U_{3} + ... + U_{n}$
How to derive the formula $S_n = (U_{n} + U_{n + 1} - 2) / 4 = \frac{(F_{3n} + F_{3n + 3} - 2)}{4}$?
 [/INST] What is being asked is to find the sum 
\begin{align}
\sum_{r=0}^{n} F_{3r}.
\end{align}
This is obtained in the following way.
\begin{align}
\sqrt{5} \, S_{n} &= \sum_{r=0}^{n} \left( \alpha^{3r} - \beta^{3r} \right) \\
&= \frac{1 - \alpha^{3n+3}}{1-\alpha^{3}} - \frac{1-\beta^{3n+3}}{1-\beta^{3}} \\
&= - \frac{1}{2\alpha} + \frac{\alpha^{3n+2}}{2} + \frac{1}{2 \beta} - \frac{\beta^{3n+2}}{2} \\
&= \frac{\sqrt{5}}{2} \left( F_{3n+2} -1 \right).
\end{align}
From this it can be stated that
\begin{align}
\sum_{r=0}^{n} F_{3r} = \frac{F_{3n+2} - 1}{2}.
\end{align}

The formulas used in this evaluation are:
\begin{align}
F_{n} = \frac{\alpha^{n} - \beta^{n}}{\alpha - \beta}
\end{align}
which is Binet's formula and
\begin{align}
\sum_{r=0}^{n} x^{r} = \frac{1 - x^{n+1}}{1-x}.
\end{align}
Other components used:
\begin{align}
1 - \alpha^{3} &= 1 - \alpha(1 + \alpha) = 1 - \alpha - (1 + \alpha) = - 2 \alpha \\
1 - \beta^{3} &= - 2 \beta \\
\alpha \beta &= -1
\end{align}
</s>