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数列与数论结合问题：a1=2,a2=3, an+1=3an-an-1 求anan+1-5是完全平方数

2014-08-09 19:32 288 查看

由
${a_{n + 1}} = 3{a_n} - {a_{n - 1}}$
,通过简单递推可以有：

$\begin{array}{l}{a_1} = 2\\{a_2} = 3\\{a_3} = 3 \times {a_2} - {a_1} = 3 \times 3 - 2 = 7\\{a_4} = 3 \times {a_3} - {a_2} = 3 \times 7 - 3 = 18\\{a_5} = 3 \times {a_4} - {a_3} = 3 \times 18 - 7 = 47\end{array}$

$\begin{array}{l}{a_1}{a_2} - 5 = 2 \times 3 - 5 = 1 = {1^2} = x{}_1^2\\{a_2}{a_3} - 5 = 3 \times 7 - 5 = 16 = {4^2} = x{}_2^2\\{a_3}{a_4} - 5 = 7 \times 18 - 5 = 121 = {11^2} = x{}_3^2\\{a_4}{a_5} - 5 = 18 \times 47 - 5 = 841 = {29^2} = x{}_4^2\end{array}$

构造一个数列
${x_n} = 3{x_{n - 1}} - {x_{n - 2}},{x_1} = 1,{x_2} = 4$
,根据上面的有限几项的数字递推关系有：

$\begin{array}{l}{a_1}{a_2} - 5 = 2 \times 3 - 5 = 1 = {1^2}\\{a_2}{a_3} - 5 = 3 \times 7 - 5 = 16 = {4^2}\\{a_3}{a_4} - 5 = 7 \times 18 - 5 = 121 = {11^2}\\{a_4}{a_5} - 5 = 18 \times 47 - 5 = 841 = {29^2}\end{array}$

假设从1,2,...,n下式均成立：

${a_n}{a_{n + 1}} - 5 = x_n^2$
(0)

那么

$\begin{array}{c}{a_{n + 1}}{a_{n + 2}} - 5 = (3{a_n} - {a_{n - 2}})(3{a_{n + 1}} - {a_n}) - 5\\ = 9{a_n}{a_{n + 1}} + {a_{n - 1}}{a_n} - 3a_n^2 - 3{a_{n - 1}}{a_{n + 1}} - 5\end{array}$
(1)

由数列递推关系
${a_{n + 1}} = 3{a_n} - {a_{n - 1}}$
,两边同时乘以
$a_n$
,于是有:

${a_n}{a_{n + 1}} = 3a_n^2 - {a_n}{a_{n - 1}} \Rightarrow 3a_n^2 = {a_n}{a_{n + 1}} + {a_n}{a_{n - 1}}$
(2)

对于数列
$a_n$
,构造一个函数
$f(k) = {a_{k - 1}}{a_{k + 1}} - a_k^2$

那么：

$\begin{array}{c}f(k + 1) - f(k) = {a_k}{a_{k + 2}} - a_{k + 1}^2 - ({a_{k - 1}}{a_{k + 1}} - a_k^2)\\ = {a_k}{a_{k + 2}} + a_k^2 - (a_{k + 1}^2 + {a_{k - 1}}{a_{k + 1}})\\ = {a_k}({a_{k + 2}} + {a_k}) - {a_{k + 1}}a({a_{k + 1}} + {a_{k - 1}})\\ = {a_k} \times 3{a_{k + 1}} - {a_{k + 1}} \times 3{a_k}\\ = 0\end{array}$

也就是说：

$f(k + 1) = f(k) = f(k - 1) = \cdots = f(3) = f(2)$

$\begin{array}{l}f(3) = {a_2}{a_4} - a_3^2 = 3 \times 18 - {7^2} = 5\\f(2) = {a_1}{a_3} - a_2^2 = 2 \times 7 - {3^2} = 5\end{array}$

于是：

$f(k) = {a_{k - 1}}{a_{k + 1}} - a_k^2 = 5$
(3)

对于构造的数列
$x_n$
, 由于其递推关系与
$a_n$
完全相同，同理可以证明齐存在下列关系：

${x_{k - 1}}{x_{k + 1}} - x_k^2 = 5$
(4)

$3x_n^2 = {x_n}{x_{n + 1}} + {x_n}{x_{n - 1}}$
(5)

将(2)、 (3)代入(1)式有：

$\begin{array}{c}{a_{n + 1}}{a_{n + 2}} - 5 = 9{a_n}{a_{n + 1}} + {a_{n - 1}}{a_n} - 3a_n^2 - 3{a_{n - 1}}{a_{n + 1}} - 5\\ = 9{a_n}{a_{n + 1}} + {a_{n - 1}}{a_n} - 3a_n^2 - 3(a_n^2 + 5) - 5\\ = 9{a_n}{a_{n + 1}} + {a_{n - 1}}{a_n} - 6a_n^2 - 20\\ = 9{a_n}{a_{n + 1}} + {a_{n - 1}}{a_n} - 2({a_n}{a_{n + 1}} + {a_{n - 1}}{a_n}) - 20\\ = 7{a_n}{a_{n + 1}} - {a_{n - 1}}{a_n} - 20\end{array}$

根据公式(0)的假定,有：

$\begin{array}{c}{a_{n + 1}}{a_{n + 2}} - {5_n} = 7{a_n}{a_{n + 1}} - {a_{n - 1}}{a_n} - 20\\ = 7(x_n^2 + 5) - (x_{n - 1}^2 + 5) - 20\\ = 7x_n^2 - x_{n - 1}^2 - 10\\ = 9x_n^2 - x_{n - 1}^2 - 2x_n^2 - 10\\ = (3{x_n} - {x_{n - 1}})(3{x_n} + {x_{n - 1}}) - 2x_n^2 - 10\\ = {x_{n + 1}} \times (3{x_n} - {x_{n - 1}} + 2{x_{n - 1}}) - 2x_n^2 - 10\\ = {x_{n + 1}} \times ({x_{n + 1}} + 2{x_{n - 1}}) - 2x_n^2 - 10\\ = x_{n + 1}^2 + 2{x_{n - 1}}{x_{n + 1}} - 2x_n^2 - 10\\ = x_{n + 1}^2 + 2({x_{n - 1}}{x_{n + 1}} - x_n^2 - 5)\\ = x_{n + 1}^2\end{array}$

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