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Using Fibonacci...

I am Proving: $$f_3 + f_6 + \cdots + f_{3n} = \frac12(f_{3n+2}-1) $$

I did the assumption of $f_1$ which gave $\mathrm{LHS}=2=\mathrm{RHS}$.

For the second part where it is $n+1$ I am having problem adding the RHS: $$f_3 + f_6 + \cdots + f_{3n}+ f_{3(n+1)} = \frac12(f_{3(n+1)+2}-1) $$

Here is the problem as I have no knowledge of how to make the function into the previous: $$\mathrm{RHS} = \frac12(f_{3n+2}-1)+f_{3(n+1)} $$

Thanks in advance... Also, if anyone got any information on properties for functions would be greatly appreciated.

Edit: Aww I understand now because in Fibonacci we can see that F(0) + f(1) = f(2) so in that perspective you can add them like that. ^^ Thank you guys... btw, it is not letting me up vote, Mark good answer, or comment

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You probably need to register in order to do more than edit. – Barry Cipra Nov 22 '13 at 23:28
up vote 1 down vote accepted

$$\begin{align} f_3+f_6+\cdots+f_{3n}+f_{3n+3}&={1\over2}\left(f_{3n+2}-1\right)+f_{3n+3}\\ &={1\over2}\left(f_{3n+2}+f_{3n+3}+f_{3n+3}-1\right)\\ &={1\over2}\left(f_{3n+4}+f_{3n+3}-1\right)\\ &={1\over2}\left(f_{3n+5}-1\right)\\ \end{align}$$

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Thank you, the step by step made me figure out that your using the fibonacci rule of F(0) + F(1) = F(2) – user110758 Nov 22 '13 at 23:37

The RHS should be $$RHS = \frac12(f_{3n+2}-1)+f_{3n+3}.$$ Observe that $\frac12 f_{3n+2}+f_{3n+3}=\frac12(f_{3n+2}+f_{3n+3})+\frac12f_{3n+3}.$ Can you finish it now?

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thanks, I just wasn't understanding on how to add them but I get it now! thanks ^^ – user110758 Nov 22 '13 at 23:38

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