smc

We compute $f(n+1)$ and $f(n-1)$, while pulling one copy of the exponential part outside: $f(n+1)=\frac{5+3\sqrt{5}}{10}\cdot \frac{1+\sqrt{5}}{2}\cdot {\left(\frac{1+\sqrt{5}}{2}\right)}^n+\frac{5-3\sqrt{5}}{10}\cdot \frac{1-\sqrt{5}}{2}\cdot {\left(\frac{1-\sqrt{5}}{2}\right)}^n$ $f(n+1)=\frac{20+8\sqrt{5}}{20}{\left(\frac{1+\sqrt{5}}{2}\right)}^n+\frac{20-8\sqrt{5}}{20}{\left(\frac{1-\sqrt{5}}{2}\right)}^n$ $f(n-1)=\frac{5+3\sqrt{5}}{10}\cdot \frac{2}{1+\sqrt{5}}\cdot {\left(\frac{1+\sqrt{5}}{2}\right)}^n+\frac{5-3\sqrt{5}}{10}\cdot \frac{2}{1-\sqrt{5}}\cdot {\left(\frac{1-\sqrt{5}}{2}\right)}^n$ $f(n-1)=\frac{(5+3\sqrt{5})(1-\sqrt{5})}{5(1+\sqrt{5})(1-\sqrt{5})}{\left(\frac{1+\sqrt{5}}{2}\right)}^n+\frac{(5-3\sqrt{5})(1+\sqrt{5})}{5(1-\sqrt{5})(1+\sqrt{5})}{\left(\frac{1-\sqrt{5}}{2}\right)}^n$ $f(n-1)=\frac{-10-2\sqrt{5}}{5(-4)}{\left(\frac{1+\sqrt{5}}{2}\right)}^n+\frac{-10+2\sqrt{5}}{5(-4)}{\left(\frac{1-\sqrt{5}}{2}\right)}^n$ $f(n-1)=\frac{10+2\sqrt{5}}{20}{\left(\frac{1+\sqrt{5}}{2}\right)}^n+\frac{10-2\sqrt{5}}{20}{\left(\frac{1-\sqrt{5}}{2}\right)}^n$ Computing $f(n+1)-f(n-1)$ gives: $f(n+1)-f(n-1)=\frac{10+6\sqrt{5}}{20}{\left(\frac{1+\sqrt{5}}{2}\right)}^n+\frac{10-6\sqrt{5}}{20}{\left(\frac{1-\sqrt{5}}{2}\right)}^n$ $f(n+1)-f(n-1)=\frac{5+3\sqrt{5}}{10}{\left(\frac{1+\sqrt{5}}{2}\right)}^n+\frac{5-3\sqrt{5}}{10}{\left(\frac{1-\sqrt{5}}{2}\right)}^n$ $f(n+1)-f(n-1)=f(n)$ Thus, the answer is $\boxed{f(n)}$.