# Prove for Fibonacci numbers

Define $$b_n$$ by $$b_1 = 1$$ and $$b_n = a_{n+1} + a_{n-1}$$ for $$n>2$$ , where $$a_n$$ is the $$n$$ th Fibonacci number.

Prove that $$a_{2n} = {a_n}{b_n}$$

I have tried induction in the question but I am unable get to an answer.

I have tried $$a_{2n+2} = {a_n}{a_{n+1}} + {a_n}{a_{n-1}} + a_{2n+1}$$

• I'm sorry... what? You may want to read the MathJax tutorial. Commented Mar 5, 2019 at 16:11
• I have edited it now. Actually, I am not versed in MathJax language. Commented Mar 5, 2019 at 16:14
• Hi Krushna, welcome to MSE! What have you tried? This is not a forum where people just do your HW for you. Please see math.meta.stackexchange.com/questions/9959/… Commented Mar 5, 2019 at 16:14
• what the value of $b_2$? inst the segond part of definition of b aplies to n > 1?
– cand
Commented Mar 5, 2019 at 16:15
• use induction, and always show your process(no matter even if they are wrong) otherwise your question won't get views and you'll be downvoted
– user427802
Commented Mar 5, 2019 at 16:15

## 2 Answers

in this answer, I’ll write $$F_n$$ and $$F_{n-1}+F_{n+1}$$ instead of $$a_n$$ and $$b_n$$, respectively, as that’s more intuitive

Use induction:

First of all, it’s clear that $$F_2\cdot(F_1+F_3)=1\cdot(1+2)=3=F_4$$.

Now assume that $$F_{2n}=F_n\cdot(F_{n-1}+F_{n+1})$$.

Then $$F_{2n+2}=F_{2n}+F_{2n+1}$$ $$=(F_n\cdot F_{n-1}+F_n\cdot F_{n+1})+(F_n^2+F_{n+1}^2)$$ $$=F_n\cdot F_{n-1}+F_n\cdot F_{n+1}+F_n\cdot(F_{n+1}-F_{n-1})+F_{n+1}\cdot(F_{n+2}-F_n)$$ $$=F_n\cdot F_{n+1} + F_{n+1}\cdot F_{n+2}$$

Note: here we use the fact that $$F_{2n+1}=F_n^2+F_{n+1}^2$$. In case you didn’t know that, it already has an answer here.

Refer here