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.
原文：https://www.cnblogs.com/mathstudy/p/8759087.html