# Fibonacci sequences

I have the following:

$$f_3+f_6 + \dots+f_{3n} = \frac 12 (f_{3n+2}-1)$$

for $f_0=0$ and $f_1=1$

When I calculate $n\ge2$ and $f_n= f_{n-1}+f_{n-2}$, I get: LHS = 8 while RHS = 10.

LHS $$f_6 =f_5+f_4 \\ f_5 = f_4+f_3 \\ f_4 = f_3 + f_2 \\ f_3=f_2+f_1 \\ f_2 = f_1 + f_0$$

and so :

$$f_2=1 \\ f_3=2 \\ f_4 = 3 \\ f_5 = 5 \\ f_6 = 8$$

RHS $$\frac 12(f_8-1) \\ \equiv \frac 12( (f_7+f_6)-1) \\ \equiv \frac 12( (f_6+f_5)+f_6-1) \\ \equiv \frac 12( (8+5)+8-1) \\ \equiv \frac 12(20) = 10$$

and so the RHS $\neq$ LHS. So how is it the basis is false when I'm asked to prove for all integers? Or is there something I missed?

• Validate using proofwiki.org/wiki/Euler-Binet_Formula Nov 18, 2013 at 16:19
• On the left hand side, you only sum $f_{3k}$, so $f_3+f_6 = 2+8 = 10 = RHS$. Nov 18, 2013 at 16:24
• Yes...you're right.. I can't believe I missed that Nov 18, 2013 at 16:28

Don't forget that your left-hand side should be $$f_3+f_6=2+8=10.$$