Question #228202

2fn-f(n-2) = fn+1 for n>3


1
Expert's answer
2021-08-23T17:48:00-0400

The recursive definition for generating Fibonacci numbers and the Fibonacci sequence is:


fn=fn1+fn2 where n3f_n = f_{n-1} + f_{n-2} \ where\ n\geq3

Then


fn+1=fn+fn1f_{n+1} = f_{n} + f_{n-1}

fn1=fn+1fnf_{n-1}=f_{n+1}-f_{n}

Substitute


fn=fn1+fn2=fn+1fn+fn2f_n = f_{n-1} + f_{n-2}=f_{n+1}-f_{n}+ f_{n-2}

Therefore


2fnfn2=fn+1,n3, True2f_n-f_{n-2} =f_{n+1}, n\geq3,\ True


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS