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初中数学八年级下册（648题）
证明：\({a_{n + 2}} = {a_n} + {a_{n + 1}}\)．
知识点：第十六章 二次根式
参考答案：即证\(\frac{1}{{\sqrt 5 }}[{(\frac{{1 + \sqrt 5 }}{2})^{n + 2}} - {(\frac{{1 - \sqrt 5 }}{2})^{n + 2}}]
\)\(= \frac{1}{{\sqrt 5 }}[{(\frac{{1 + \sqrt 5 }}{2})^n} - {(\frac{{1 - \sqrt 5 }}{2})^n}] +
\)\( \frac{1}{{\sqrt 5 }}[{(\frac{{1 + \sqrt 5 }}{2})^{n + 1}} - {(\frac{{1 - \sqrt 5 }}{2})^{n + 1}}]\)
即\({(\frac{{1 + \sqrt 5 }}{2})^{n + 2}} - {(\frac{{1 - \sqrt 5 }}{2})^{n + 2}}
\)\(= {(\frac{{1 + \sqrt 5 }}{2})^n} - {(\frac{{1 - \sqrt 5 }}{2})^n} + {(\frac{{1 + \sqrt 5 }}{2})^{n + 1}} - {(\frac{{1 - \sqrt 5 }}{2})^{n + 1}}\)，
即\({(\frac{{1 + \sqrt 5 }}{2})^n} \times {(\frac{{1 + \sqrt 5 }}{2})^2} - {(\frac{{1 - \sqrt 5 }}{2})^n} \times {(\frac{{1 - \sqrt 5 }}{2})^2}\)
\( = {(\frac{{1 + \sqrt 5 }}{2})^n} - {(\frac{{1 - \sqrt 5 }}{2})^n} +
\)\( {(\frac{{1 + \sqrt 5 }}{2})^n} \times (\frac{{1 + \sqrt 5 }}{2}) - {(\frac{{1 - \sqrt 5 }}{2})^n} \times (\frac{{1 - \sqrt 5 }}{2})\)，
即\({(\frac{{1 + \sqrt 5 }}{2})^n} \times \left( {\frac{{3 + \sqrt 5 }}{2} - 1 - \frac{{1 + \sqrt 5 }}{2}} \right) -
\)\( {(\frac{{1 - \sqrt 5 }}{2})^n} \times \left( {\frac{{3 - \sqrt 5 }}{2} - 1 - \frac{{1 - \sqrt 5 }}{2}} \right) = 0\)，
即\({(\frac{{1 + \sqrt 5 }}{2})^n} \times 0 - {(\frac{{1 - \sqrt 5 }}{2})^n} \times 0 = 0\)，得证．