Not understanding notation for a question on fibonacci sequences

I'm trying to solve a problem via induction involving fibonacci sequences, but I'm confused as to what this notation is. It's mostly the ... , is that adding from $F_1$ to $F_n$ then subtracting $F_{2n-1}$? Seems really weird... thanks for your help! And you're welcome to provide tips for the problem if you like, it's much appreciated!

$$\text{Prove, using induction, } F_0 - F_1 + F_2 - \cdots - F_{2n-1} + F_{2n} = F_{2n-1} -1; \qquad (F_0=0).$$

• I'm writing an answer, but is $F_{1}=1$? Commented Apr 6, 2017 at 22:25
• The first element of a Fibonacci sequence Commented Apr 6, 2017 at 22:27

2 Answers

It means $$\sum\limits_{k=0}^{2n}(-1)^k F_k$$ i.e. it's the sum but with every other changing sign. For example with $n = 4$, it's $$F_0 - F_1+F_2-F_3+F_4-F_5+F_6-F_7+F_8 = F_7 - 1.$$

We are attempting to prove the following proposition by induction:

$$P_{n}: \sum_{r=0}^{2n}(-1)^rF_{r}=F_{2n-1}-1,\quad n\in\mathbb{Z^+},\quad F_{0}=0,F_{1}=1$$

So first we must show it's true for $n=1:$

$P_{1} : \sum_{r=0}^{2}(-1)^rF_{r}= F_{0}-F_{1}+F_{2} = 0-1+1=0 = 1 - 1 =F_{1}-1$ as required, hence $P_{1}$ is true.

Then we assume $P_{n}$ to be true, and attempt to prove $P_{n+1}$

$$P_{n+1}:\sum_{r=0}^{2(n+1)}(-1)^rF_{r}= \sum_{r=0}^{2n}(-1)^rF_{r} + (-1)^{2n+1}F_{2n+1}+(-1)^{2n+2}F_{2n+2}$$ $$= \underbrace{F_{2n-1}-1}_{\text{using our assumption}}-F_{2n+1}+F_{2n+2}$$

Note that $F_{2n-1}+F_{2n}=F_{2n+1}\implies F_{2n-1}=F_{2n+1}-F_{2n}$

$F_{2n}+F_{2n+1}=F_{2n+2}\implies F_{2n}=F_{2n+2}-F_{2n+1}$

Then $F_{2n-1}=F_{2n+1}-(F_{2n+2}-F_{2n+1})=2F_{2n+1}-F_{2n+2}$

$$P_{n+1}: F_{2n-1}-1-F_{2n+1}+F_{2n+2} = 2F_{2n+1}-F_{2n+2}-1-F_{2n+1}-F_{2n+2}$$

$$=F_{2n+1}-1$$ as required, hence $P_{n}$ holds $\forall n\in\mathbb{Z^+}$