MT【129】常数变易法

已知数列\(\{x_n\}\)满足$$x_{n+1}=\left(\dfrac 2{n^2}+\dfrac 3n+1\right)x_n+n+1,n\in\mathbf N^*,$$且\(x_1=3\)，求数列\(\{x_n\}\)的通项公式．

解答:
根据题意，有$$x_{n+1}=\dfrac{(n+1)(n+2)}{n^{2}x_n+n+1,$$于是$$\dfrac{x_{n+1}}{(n+1)}2(n+2)}=\dfrac{x_n}{n^2(n+1)}+\dfrac{1}{(n+1)(n+2)},$$ 进而可得$$\dfrac{x_{n+1}}{(n+1)^{2(n+2)}+\dfrac{1}{n+2}=\dfrac{x_n}{n}2(n+1)}+\dfrac{1}{n+1},$$ 因此$$\dfrac{x_n}{n^{2(n+1)}+\dfrac{1}{n+1}=\dfrac{x_{n-1}}{(n-1)}2\cdot n}+\dfrac{1}{n}=\cdots =\dfrac{x_1}{2}+\dfrac 12=2,$$所以\(x_n=n^2(2n+1),n\in\mathbf N^*\)．
评:这里除去的这一项\((n+1)^2(n+2)\)是由常数变易法得来的.