if f(n+2)=2f(n)+f(n+1) and f(1)=1 and f(2)=1 , what is f(8)-f(7) ?

Question
Functions
asked 2021-03-12
if \(\displaystyle{f{{\left({n}+{2}\right)}}}={2}{f{{\left({n}\right)}}}+{f{{\left({n}+{1}\right)}}}\) and f(1)=1 and f(2)=1 , what is f(8)-f(7) ?

Answers (1)

2021-03-13
Given that f(n+2)=2f(n)+f(n+1) with f(1)=1and f(2)=1. Start with putting n=6 in the above we have f(8)-f(7)=2f(6)
=2(2f(4)+f(5)) =2(2f(4)+2f(3)+f(4)) =2(3f(4)+2f(3)) =2(3f(2f(2)+f(3))+2f(3)) =2(6f(2)+3f(3)+2f(3)) =2(6f(2)+5f(3)) =2(6f(2)+5(2f(1)+f(2))) =2(6f(2)+10f(1)+5f(2)) =2(11f(2)+10f(1)) =2(11+10) =\(\displaystyle{2}\cdot{21}\) =42
0

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