# Let $F_n$ be the Fibonacci number, with $F_0 = 1, F_1 = 1$. Prove the identity $2F_n = F_{n+1} + F_{n−2}$ by using generating functions.

Let Fn be the Fibonacci number, with $$F_0 = 1, F_1 = 1$$. Prove the identity $$2F_n = F_{n+1} + F_{n−2}$$ by using generating functions.

I know the generating function for the shifted Fibonacci Sequence, $$F_n$$ with $$F_0=1$$ and $$F_1=1$$ but not sure how to incorporate the $$2$$.

• math.meta.stackexchange.com/questions/5020/…
– user645636
Jan 25, 2020 at 15:57
• Can you tell us the generating function then? That said, this question is far easier to solve from the basic definition.
– user694818
Jan 25, 2020 at 18:06

## 1 Answer

You may prove it by using the direct formula for Fibonacci numbers $$F_k=\frac{a^k-b^k}{\sqrt{5}}, a+b=1, ab=-1,a^2=a+1, b^2=b+1$$, Then $$F_{n+1}+F_{n-2}= \frac{a^{n+1}-b^{n+1}}{\sqrt{5}}+\frac{a^{n-2}-b^{n-2}}{\sqrt{5}}$$ $$=\frac{1}{\sqrt{5}}\left(a^n(a+\frac{1}{a^2})-b^n(b+\frac{1}{b^2}) \right)$$ note that $$a+1/a^2=a+b^2=a+b+1=2$$, then $$F_{n+1}+F_{n-2}= \frac{2}{\sqrt{5}}[a^n-b^n]=2 F_n$$

• If you're going to avoid generating functions, you may as well prove this from $F_{n+1}=F_n+F_{n-1}$ and $F_n=F_{n-1}+F_{n-2}$ and middle school algebra.
– user694818
Jan 25, 2020 at 18:10
• Yes, you are right, but I like it this way. Jan 25, 2020 at 18:14