回復 1# thankyou 的帖子
第1題
這題不用數學歸納法比較容易
第2題
\(\begin{align}
& {{a}_{n+1}}+{{a}_{n}} \\
& =\frac{1}{\sqrt{5}}\left[ {{\left( \frac{1+\sqrt{5}}{2} \right)}^{n+1}}-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n+1}}+{{\left( \frac{1+\sqrt{5}}{2} \right)}^{n}}-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n}} \right] \\
& =\frac{1}{\sqrt{5}}\left[ {{\left( \frac{1+\sqrt{5}}{2} \right)}^{n}}\left( \frac{1+\sqrt{5}}{2}+1 \right)-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n}}\left( \frac{1-\sqrt{5}}{2}+1 \right) \right] \\
& =\frac{1}{\sqrt{5}}\left[ {{\left( \frac{1+\sqrt{5}}{2} \right)}^{n}}{{\left( \frac{1+\sqrt{5}}{2} \right)}^{2}}-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n}}{{\left( \frac{1-\sqrt{5}}{2} \right)}^{2}} \right] \\
& =\frac{1}{\sqrt{5}}\left[ {{\left( \frac{1+\sqrt{5}}{2} \right)}^{n+2}}-{{\left( \frac{1-\sqrt{5}}{2} \right)}^{n+2}} \right] \\
& ={{a}_{n+2}} \\
\end{align}\)
後面就不做了
