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Here is what I have so far for proof by induction:

Base case:

Suppose n = 3

$$\sum_{i=1}^{n}(F_{2n-1} - F_{2n}) = 1 - F_{2n-1}$$ $$\sum_{i=1}^{3}(F_{2n-1} - F_{2n}) = 1 - F_{2*3-1}$$ $$F_{1} - F_{2} + F_{3} - F_{4} + F_{5} - F_{6} = 1 - F_{5}$$ $$1 - 1 + 2 - 3 + 5 - 8 = 1 - 5$$ $$-4 = -4$$

Inductive hypothesis:

$$\sum_{i=1}^{k}(F_{2k-1} - F_{2k}) = 1 - F_{2k-1} \forall k \geq 1$$

Inductive step:

$$\textrm{We must prove that} \ \sum_{i=1}^{k+1}(F_{2k+1} - F_{2k+2}) = 1 - F_{2k+1}$$

This is where I get stuck. I am also unsure whether the equation in the inductive step is actually correct or not. I know I have to sub in k+1 for k, but when I try a base case with the new equation to see if it is still true, I can't actually make it work out. I have been at this question for over an hour now so maybe my brain is just fried. Have I been going at this question wrong?

Pizza Pizza
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  • Seems you have a mistake in your inductive step $F_{2(k+1)-1}=F_{2k+1}\neq F_{2k}$. – kingW3 Jun 10 '17 at 17:56
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    In your question's title you have $;F_{2n}=1-F_{2n-1};$ . This can't be right...unless I misunderstood something and $;F_n;$ is not the Fibonacci Sequence – DonAntonio Jun 10 '17 at 18:00
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    See https://math.stackexchange.com/questions/787341/summation-of-fibonacci-numbers-f-n-with-n-odd-vs-even. – lhf Jun 10 '17 at 18:05
  • I have edited the title to fix what I think was a mistake. If this is not what you meant, feel free to edit it yourself. – Wojowu Jun 10 '17 at 18:35
  • I realize where I made a mistake in my inductive step! I have edited my post accordingly. Also, the title is now correct. – Pizza Pizza Jun 10 '17 at 19:19

2 Answers2

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Note that the sum "telescopes" using the Fibonacci recursion $F_{2n}=F_{2n-1}+F_{2n-2}$, so that $$ \sum_{i=1}^n(F_{2i-1}-F_{2i})=-\left(\sum_{i=0}^{n-1}F_{2i}\right). $$ Thus it suffices to show $$ \sum_{i=0}^{n-1}F_{2i}=F_{2n-1}-1; $$ this is much easier to induct with, as if $$ F_0+\dots+F_{2n-2}=F_{2n-1}-1, $$ then $$ F_0+\dots+F_{2n-2}+F_{2n}=F_{2n-1}-1+F_{2n}=F_{2n+1}-1 $$

TomGrubb
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Why do you have $n=3$ as your base case? Why not just $n=1$?

Also, make sure to use the running index $i$ as your variable in the formula inside the summation (you are using $n$ and $k$, which is not right!)

So I would do:

Claim: for all $n \ge 1$:

$$\sum_{i=1}^{n}(F_{2i-1}-F_{2i})=1-F_{2n-1}$$

Base: $n=1$

$$F_1-F_2=1-1=1-F_1$$ Check!

Step:

Inductive hypothesis: Suppose that for some arbitrary $k$:

$$\sum_{i=1}^{k}(F_{2i-1}-F_{2i})=1-F_{2k-1}$$

now consider $k+1$:

$$\sum_{i=1}^{k+1}(F_{2i-1}-F_{2i})=$$

$$\sum_{i=1}^{k}(F_{2i-1}-F_{2i})+F_{2(k+1)-1}-F_{2(k+1)}=\text{ (Inductive Hypothesis)}$$

$$1-F_{2k-1}+F_{2k+1}-F_{2k+2}=$$

$$1-F_{2k-1}+F_{2k-1}+F_{2k}-F_{2k}-F_{2k+1}=$$

$$1-F_{2(k+1)-1}$$

Check!

Bram28
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