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February 22nd, 2017, 05:32 PM   #11
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In fact, it is a recursive problem.
Known $\displaystyle f(0)=1,f(1)=1,f(n)=f(n-1)+f(n-2)$
One obtains $\displaystyle f(n)=(\frac{\sqrt{5}+1}{2})^n+(\frac{\sqrt{5}-1}{2})^n$
Similarly

Known $\displaystyle f(0)=2,f(1)=6,f(n)=3f(n-1)-f(n-2)$
One obtains $\displaystyle f(n)=(\frac{3+\sqrt{5}}{2})^n+(\frac{3-\sqrt{5}}{2})^n=(\frac{\sqrt{5}+1}{2})^{2n}+(\frac {\sqrt{5}-1}{2})^{2n}$
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