<s>[INST] Fibonacci sequence proof Prove the following:
$$f_3+f_6+...f_{3n}= \frac 12(f_{3n+2}-1) \\ $$
For $n \ge 2$
Well I got the basis out of the way, so now I need to use induction: So that $P(k) \rightarrow P(k+1)$ for some integer $k \ge 2$
So, here are my first steps:
$$ \begin{align} & \frac 12(f_{3k+2}-1) + f_{3k+3} = \\ 
   & = \frac 12(f_{3k+2}-1) + f_{3k+1} + f_{3k} +  f_{3k+1} \\
   & = \frac 12(f_{3k+2}-1) + f_{3k+1} + \frac 12(f_{3k+2}-1) +  f_{3k+1} \\
   & = f_{3k+2}-1 + 2 \cdot f_{3k+1} 
   \end{align} $$
And the fun stops around here. I don't see how to get to the conclusion: $\frac 12(f_{3k+5}-1) \\ $. Any help from this point would be great.
 [/INST] define 
$$B_n = \sum_{k=1}^{3n} f_k$$
LEMMA $$B_n = f_{3n+2} - 1$$
PROOF by induction. 
(A) for $n=1$ we have $f_1+f_2+f_3 = 1+1+2=4 = 5-1=f_5-1$
(B) suppose true for $n$, then:
$$B_{n+1} = B_n +f_{3n+1} +f_{3n+2}+f_{3n+3} \\
= f_{3n+2} - 1 +f_{3n+1} +f_{3n+2}+f_{3n+3} \\
= (f_{3n+1} +f_{3n+2}) +(f_{3n+2}+f_{3n+3})-1\\
= f_{3n+3} +f_{3n+4}  -1\\
= f_{3n+5} -1\\
= f_{3(n+1)+2} -1 \\
$$
qed
now define $S_n = \sum_{k=1}^n f_{3k}$
clearly $B_n=2S_n$ since $f_1+f_2+f_3 = 2f_3$ etc.
and the result follows.
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