复杂递推求通项,归纳换元造零常

复杂递推求通项,归纳换元造零常。

【例题】已知各项都为正数的数列\(\{ {a_n}\} \)满足\({a_{n + 2}} = 2{a_{n + 1}} + 3{a_n}\).

(1)证明:数列\(\{ {a_n} + {a_{n + 1}}\} \)为等比数列;

(2)若\({a_1} = \dfrac{1}{2}\),\({a_2} = \dfrac{3}{2}\),求\(\{ {a_n}\} \)的通项公式.

【解析】(1)\({a_{n + 2}} = 2{a_{n + 1}} + 3{a_n}\)

\(\Leftrightarrow {a_{n + 2}} + {a_{n + 1}} = 3{a_{n + 1}} + 3{a_n} = 3({a_{n + 1}} + {a_n})\)

\(\because {a_n}>0\),\(\therefore \dfrac{{{a_{n + 1}} + {a_{n + 2}}}}{{{a_n} + {a_{n + 1}}}} = 3\)(非零常数)

故数列\(\{ {a_n} + {a_{n + 1}}\} \)为等比数列

(2)由(1)知道数列\(\{ {a_n} + {a_{n + 1}}\} \)为等比数列,且首项为\({a_1} + {a_2} = 2\),公比为\(3\)

故\({a_n} + {a_{n + 1}} = 2 \times {3^{n – 1}}\)(*)

【\({a_1} = \dfrac{1}{2}\),\({a_2} = \dfrac{3}{2}\),\({a_3} = \dfrac{9}{2}\),\({a_4} = \dfrac{{27}}{2}\),猜想\({a_n} = \dfrac{1}{2} \times {3^{n – 1}}\)】

令\({b_n} = {a_n} – \dfrac{1}{2} \times {3^{n – 1}}\),则\({a_n} = {b_n} + \dfrac{1}{2} \times {3^{n – 1}}\)

代入(*)得\({b_n} + \dfrac{1}{2} \times {3^{n – 1}} + {b_{n + 1}} + \dfrac{1}{2} \times {3^n} = 2 \times {3^{n – 1}}\),整理得\({b_{n + 1}} = – {b_n}\)

由于\({b_1} = {a_1} – \dfrac{1}{2} = 0\),故\({b_n} = 0\),从而\({a_n} = \dfrac{1}{2} \times {3^{n – 1}}\)

【反思】本题也可以用待定系数法求解,指数式可以分成两个指数式,要注意规律性:

设​\( a_{n+1}+a_n=2\times3^{n-1}可变形为a_{n+1}+\lambda\times3^{n}=-(a_n+\lambda\times3^{n-1}) \)

原创文章,作者:leopold,如若转载,请注明出处:https://www.math211.com/2021/02/10/82/

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