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高中数学选择性必修 第二册(381题)


已知数列\(\left\{ {{a_n}} \right\}\) 中,\({a_1} = {a_2} = 1\),且 \({a_{n + 2}} = {a_{n + 1}} + 2{a_n}\) .记 \({b_n} = {a_{n + 1}} + {a_n}\)

求证:数列 \(\left\{ {{b_n}} \right\}\) 是等比数列;



知识点:第四章 数列


参考答案:∵\(\frac{{{b_{n + 1}}}}{{{b_n}}} = \frac{{{a_{n + 2}} + {a_{n + 1}}}}{{{a_{n + 1}} + {a_n}}} = \frac{{{a_{n + 1}} + 2{a_n} + {a_{n + 1}}}}{{{a_{n + 1}} + {a_n}}} = \frac{{2\left( {{a_{n + 1}} + {a_n}} \right)}}{{{a_{n + 1}} + {a_n}}} = 2\),且 \({b_1} = {a_1} + {a_2} = 2 \ne 0\),∴ \(\left\{ {{b_n}} \right\}\) 是以2为首项,2为公比的等比数列;

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