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I'm stuck on proving this statement below due to that I can't seem to find a base case that is true:

Prove that $$\sum_{i=0}^{2n}(-1)^if(i) = f(0) - f(1) + f(2) - \cdots - f(2n-1) + f(2n) = f(2n-1)-1,$$ where $f(i)$ is the $i$th Fibonacci number.

I believe simple induction should work but I am unsure if I have to use Strong Induction.

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marked as duplicate by lulu, Namaste discrete-mathematics Mar 21 '17 at 0:08

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  • $\begingroup$ For what $n$ do you need to prove this? $\endgroup$ – Mastrem Mar 20 '17 at 20:09
  • $\begingroup$ The sum looks like it should be $\sum (-1)^iF(i)$ not $F(2i)$. $\endgroup$ – Mark Fischler Mar 20 '17 at 20:11
  • $\begingroup$ why you are "unsure" about using weak or strong induction? Just try the weak induction approach, if it gets complicate try the other. So you dont need to be unsure! $\endgroup$ – Masacroso Mar 20 '17 at 20:12
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For $n+1$ we have

$$\sum_{i=0}^{2(n+1)}(-1)^if(i)=\sum_{i=0}^{2n}(-1)^if(i)-f(2n+1)+f(2n+2)=\\ =f(2n-1)-1-f(2n+1)+f(2n+2)$$

but

$$f(2n+2)=f(2n+1)+f(2n),\\ f(2n+1)=f(2n)+f(2n-1)$$ so,

$$f(2n-1)-f(2n+1)+f(2n+2)=f(2n+1)$$

and then

$$\sum_{i=0}^{2(n+1)}(-1)^if(i)=f(2n+1)-1=f[2(n+1)-1]-1$$

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