$7 \cdot 98$ Let $f(n)=\frac{5+3 \sqrt{5}}{10}\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\frac{5-3 \sqrt{5}}{10} \cdot\left(\frac{1-\sqrt{5}}{2}\right)^{n}$, then $f(n+1)-f(n-1)$, expressed in terms of $f(n)$ is
(A) $\frac{1}{2} f(n)$.
(B) $f(n)$.
(C) $2 f(n)+1$.
(D) $f^{2}(n)$.
(E) $\frac{1}{2}\left[f^{2}(n)-1\right]$.
(15th American High School Mathematics Examination, 1964)
[Solution] Let $\frac{1+\sqrt{5}}{2}=\alpha, \frac{1-\sqrt{5}}{2}=\beta$, then
$$
\begin{aligned}
& f(n+1)-f(n-1) \\
= & \left(\frac{5+3 \sqrt{5}}{10} \alpha^{n+1}+\frac{5-3 \sqrt{5}}{10} \beta^{n+1}\right)-\left(\frac{5+3 \sqrt{5}}{10} \alpha^{n-1}+\frac{5-3 \sqrt{5}}{10} \beta^{n-1}\right) \\
= & \frac{5+3 \sqrt{5}}{10} \alpha^{n}\left(\alpha-\frac{1}{\alpha}\right)+\frac{5-3 \sqrt{5}}{10} \beta^{n}\left(\beta-\frac{1}{\beta}\right) \\
= & \frac{5+3 \sqrt{5}}{10} \alpha^{n}+\frac{5-3 \sqrt{5}}{10} \beta^{n}=f(n) .
\end{aligned}
$$

Notice that $\alpha-\frac{1}{\alpha}=1, \quad \beta-\frac{1}{\beta}=1$.
Therefore, the answer is $(B)$.