Prove that $F_1 + F_3 + ... + F_{2n-1} = F_{2n}$ when $n$ is a positive integer

Question

I'm not sure how to use the information to prove this,what is $F_{2n-1}$? and what is $F_{2n}$? Apparently I need to use induction,

Basis Step $n= 0$ $f(0)-1 = f(2\cdot 0)$ But what does that do, I'm not sure what to do here

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How do they know $F_1 = 1$ and that it $= F_2$ and for inductive how are they able to add $F_{2k+1}$

Answer 1

By induction: $$f_1+f_3+...+f_{2n-1}+f_{2n+1}=f_{2n}+f_{2n+1}=f_{2n+2}$$ and since $f_1=f_2$, we are done!

Answer 2

$f_n$ is the $n$th Fibonacci number, which is defined as follows: $f_0=0$ and $f_1=1$ and $f_{n+2}=f_{n+1}+f_n$ for every $n\in \mathbb N.$ In paricular we have $f_{2n+1}+f_{2n}=f_{2n+2}.$ The first few Fibonacci numbers are $0,1,1,2,3,5,8,13,21,34,...$

Let $S(n)$ stand for the statement $\sum_{j=1}^{j=n}f_{2j-1}=f_{2n}.$ We want to prove that $S(n)$ is true for every $n\in \mathbb N$ by induction on $n$. There are two steps:

(I). Prove that $S(1)$ is true.

(II). Prove that the following sentence,which I will call $T,$ is true:

If $n\in \mathbb N$ and $S(n)$ is true then $S(n+1)$ is true.

Note that $T,$ which needs to be proved, does not assert that $S(n)$ actually is true for any $n.$ It just says that IF $S(n)$ THEN $S(n+1).$

To prove (I): When $n=1$ we have $2n-1=1=n$ and $\sum_{j=1}^{j=n}f_{2j-1}=f_1=f_2=f_{2n}.$

To prove (II): $$S(n) \implies \sum_{j=1}^nf_{2j-1}=f_{2n} \implies$$ $$\implies \sum_{j=1}^{j=n+1}f_{2j-1}=f_{2n+1}+\sum_{j=1}^{j=n}f_{2j-1} =f_{2n+1}+f_{2n}=f_{2n+2}\implies$$ $$\implies \sum_{j=1}^{j=n+1}f_{2j-1}=f_{2(n+1)}.$$ The last line above is exactly $S(n+1)$ so we have $S(n)\implies S(n+1).$ That is, we have proved $T$.

Answer 3

$F_1=F_2=1$, then:

$F_3=F_4-F_2$

$F_5=F_6-F_4$

etc., and the sum telescopes.