document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #16537 by venugopalramana(3286) $\"\"$ $\"About$

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