<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] It is relatively easy to show that , for  $ n \geqslant 1 $, $4F_{3n} = F_{3(n+1)}-F_{3(n-1)}$(Hint: Use the definition of the Fibonacci numbers). Using this identity, we can now show that $$\sum_{k=1}^n F_{3k} = \frac{(F_{3n} + F_{3n + 3} - 2)}{4} $$
by summing up $4F_{3k}$, where k goes from $1$ to $n$.
$$\begin{align} 4 \sum_{k=1}^n F_{3k} &= \sum_{k=1}^n 4F_{3k}= \sum_{k=1}^n F_{3(k+1)}-F_{3(k-1)} \\
&= \sum_{k=1}^n F_{3(k+1)} - \sum_{k=1}^n F_{3(k-1)} \\&= \sum_{k=2}^{n+1} F_{3k} - \sum_{k=0}^{n-1} F_{3k} \\ &= F_{3n} + F_{3n + 3} + \sum_{k=2}^{n-1} F_{3k} - \sum_{k=2}^{n-1} F_{3k} -2 \\ &= F_{3n} + F_{3n + 3} - 2
\end{align}$$
Dividing everything by $4$ , we reach the desired equality.
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