<s>[INST] Proof by Induction for a $f_3 + f_6 + · · · + f_{3n} = \frac{1}{2} (f_{3n+2} - 1)$ The Fibonacci numbers are defined as follows: $f_0 = 0,\ f_1 = 1$, and for $n ≥ 2,\ f_n = f_{n−1} +f_{n−2}$. Prove that for every positive integer $n$,
$$f_3 + f_6 + \ldots + f3_n = \frac{1}{2} (f_{3n+2} - 1) $$
Here are my steps:
Basis Step:
For $n ≥ 1$
$f_{3\cdot1} = \frac{1}{2} (f_{3\cdot 1+2} - 1) $
LHS $= f_3 = 2$
RHS $= \frac{1}{2} (f_5 - 1) =  \frac{1}{2} (5 – 1) = 2$
Induction Step:
Assume that $f_3 + f_6 + \ldots + f_{3n} = \frac{1}{2} (f_{3n+2} - 1)$  for some $n ≥ 1$.
Show that $f_3 + f_6 + \ldots + f_{3n}  + f_{3(n+1)} = \frac{1}{2} (f_{3(n+1)+2} – 1)$.
$$\frac{1}{2} (f_{3n+2} – 1) = \frac{1}{2} (f_{3(n+1)+2} – 1) + f_{3(n+1)}\ \text{(Basis)}\\ \text{LHS}=\frac{1}{2}(f_{3n+5}-1)+f_{3n+3}$$
I don't know how to proceed further. I understand what I have to do, but I don't know where to start reducing it using the fib definition.
 [/INST] $$f_3 + f_6 + · · · + f_{3n} =\frac{1}{2}(f_{3n+2} - 1) $$
$$f_3+f_6 + ···+f_{3n}+f_{3(n+1)} =\frac{1}{2}(f_{3n+2} - 1)+f_{3n+3}= $$
$$=\frac{1}{2}((f_{3n+2}+f_{3n+3})+f_{3n+3}-1)=\frac{1}{2}((f_{3n+4}+f_{3n+3}) - 1)= $$
$$=\frac{1}{2}(f_{3n+5} - 1)=\frac{1}{2}(f_{3(n+1)+2} - 1) $$
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