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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?

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

1 Answer 1

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

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