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Using induction prove that: $$\sum_{i=0}^{2n} (-1)^iF_i = (F_{2n-1})-1$$

I am on the inductive step but am stuck with how to proceed.

This is what I have after the base case so far:

$$\sum_{i=0}^{2n+2} (-1)^iF_i = (F_{2n+2})-1$$ = $$(F_{2n+1})+F_{2n}-1$$ = $$(F_{2n-1})+F_{2n}+F_{2n}-1$$ I'm not sure how to simplify this out

JanoyCresva
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1 Answers1

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Sketch: For $n=1$, the LHS becomes $$ F_0-F_1+F_2=0-1+1=0 $$ and the RHS becomes $$ F_1-1=1-1=0. $$ Since the LHS and RHS are both zero, they are equal.

Now, assume that the claim is true for $n=k$. Therefore $$ \sum_{i=0}^{2k}(-1)^iF_i=F_{2k-1}-1. $$ Consider the case where $n=k+1$. In this case, the LHS is $$ \sum_{i=0}^{2(k+1)}(-1)^iF_i=F_{2k+2}-F_{2k+1}+\sum_{i=0}^{2k}(-1)^iF_i. $$ Now, use the inductive hypothesis and break up your Fibonacci numbers. Recall that your goal is to get that the RHS is $F_{2(k+1)-1}-1=F_{2k+1}-1$, so break things up to try to get towards that expression.

Michael Burr
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  • I'm confused with where the $F_{2k+1}$ came from – JanoyCresva Feb 16 '17 at 01:13
  • Look at the upper limits of the sums, the original sum goes from $i=0$ all the way to $i=2k$. On the other hand, the bigger sum goes from $i=0$ all the way to $i=2(k+1)=2k+2$. Therefore, the sum has two extra terms, when $i=2k+1$ and when $i=2k+2$. To cut down the sum to the inductive case, you need to remove the first two terms. $F_{2k+2}$ and $-F_{2k+1}$ are those two terms (after plugging in the two extra values for $i$). – Michael Burr Feb 16 '17 at 01:16
  • When $n=3$, the sum is $\sum_{i=0}^{2n}(-1)^iF_i=\sum_{i=0}^6(-1)^iF_i=(F_0-F_1+F_2-F_3+F_4-F_5+F_6)$. On the other hand, when $n=4$, the sum is $\sum_{i=0}^{2n}(-1)^iF_i=\sum_{i=0}^8(-1)^iF_i=(F_0-F_1+F_2-F_3+F_4-F_5+F_6-F_7+F_8)$. Observe that the sum when $n=4$ has the two extra terms $F_8-F_7$. – Michael Burr Feb 16 '17 at 01:19
  • A common mistake with sums is to forget that the upper limit matters. A lot of times, people are so used to sums that have one extra term each time you increase $n$, they forget to look at the upper limit. The upper limit in this case is a formula, so each time $n$ increases, you get two extra terms, not one. – Michael Burr Feb 16 '17 at 01:21
  • So, removing the extra terms from the sum is a step you have to take any time you're doing induction involving these sums? – JanoyCresva Feb 16 '17 at 01:22
  • It is a common trick in induction involving sums, but it's not the only possible trick. – Michael Burr Feb 16 '17 at 01:23
  • Ok I got it now. I just don't understand the exact way you get the two terms.I understand why they are there though. – JanoyCresva Feb 16 '17 at 01:33
  • Let us continue this discussion in chat. – JanoyCresva Feb 16 '17 at 01:42