# Fibonacci sequence: what's wrong here?

Given f1+f3+f5+...+f2n-1 = f2n.

Prove by induction.

So using the general Fibonacci sequence formula, Fn=Fn-1 + Fn-2 ,

I got f1+f3+...+f2n+1 = f2n+2
then using the formula, f1+f3+...+f2n+1 +f2n = f2n+2.
This ends up giving me, f2n + f2n does not equal f2n+2.

What's wrong here?

( I already proved the base case by the way).

## Answers and Replies

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dextercioby
Homework Helper
Hmm... Can you plug 'n+1' instead of 'n' in the equation you should assume as known ?

So plug n+1 into Fibonacci equation? Did that, and got Fn+ Fn-1.
Then used this for f2n+1 on the l.h.s. To get f2n + f2n + f2n+1 on the left hand side. But this doesn't equal f2n+1 does it?

dextercioby
Homework Helper
OK, let's take it this way

Make $n\rightarrow n+1$ in

[tex] ...+f_{2n-3} + f_{2n-1} = f_{2n} [/itex]

and work in the LHS of the new equality only. Then compare with what you have to prove (which would have to coincide with the RHS of the new equality).

For n= n + 1, I get:

...+ f2n-1 + f2n+1 = f2n+2.
So we can state that f2n + f2n+1 = f2n+2...which is true, but my issue is why does this not work when you use the fibonacci formula?

Let me give another example, like to prove that f3n is an even integer, we can apply the formula, and it works. But in the problem stated above, it does not seem to work the same way. It only works when you add F2n+1 to both sides of the equation.

Dick
Homework Helper
Given f1+f3+f5+...+f2n-1 = f2n.

Prove by induction.

So using the general Fibonacci sequence formula, Fn=Fn-1 + Fn-2 ,

I got f1+f3+...+f2n+1 = f2n+2
then using the formula, f1+f3+...+f2n+1 +f2n = f2n+2.
This ends up giving me, f2n + f2n does not equal f2n+2.

What's wrong here?

( I already proved the base case by the way).
Ok, so the induction step is f_1+f_3+...+f_{2n-1}+f_{2n+1}=f_{2n+2}, given that f_1+f_3+...+f_{2n-1}=f_{2n}. If you put the given in doesn't that give you f_{2n}+f_{2n+1}=f_{2n+2}? I'm really confused about what you are confused about.