SOLUTION: The Fibonacci numbers 1,1,2,3,5,8.... are defined by F0=F1 and Fn=Fn-1+Fn-2 for n<=2. Show that Fn<=(7/4)^n.

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Question 29759: The Fibonacci numbers 1,1,2,3,5,8.... are defined by F0=F1 and Fn=Fn-1+Fn-2 for n<=2. Show that Fn<=(7/4)^n.
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
The Fibonacci numbers 1,1,2,3,5,8.... are defined by F0=F1 and Fn=Fn-1+Fn-2 for n<=2. Show that Fn<=(7/4)^n.
SEE BELOW
The Fibonacci numbers 1,1,2,3,5,8....I THINK THEY START WITH 0,1,1,2,3,5,8....
are defined by F0=F1 ...THEY ARE DEFINED BY F0=0 AND F1=1....
and Fn=Fn-1+Fn-2 for n<=2.........NO...FOR N>=2
Show that Fn<=(7/4)^n.
USING THOSE CHANGES WE NEED TO SHOW..Fn<=(7/4)^(n-1).AS WE ARE STARTING WITH 0 INSTEAD OF 1.
THIS SEQUENCE IS DETERMINED BY THE GENERAL EQN.......
FN=F(N-1)+F(N-2)..IN SUCH A CASE FROM ALGEBRA ,WE GET..THAT..
FN=(A^N-B^N)/(A-B),WHERE A AND B ARE ROOTS OF THE QUADRATIC.....
X^2-X-1=0...(IF YOU WANT TO KNOW ,HOW WE GET THIS IN ALGEBRA,PLEASE COME BACK)
SO A=(1+SQRT 5)/2 AND B = (1-SQRT 5)/2..
HENCE FN=A^(N-1)+N*A^(N-2)*B+.... =(7/4)^(N-1)