已知数列({x_n})满足$$x_{n+1}=left(dfrac 2{n^2}+dfrac 3n+1
ight)x_n+n+1,ninmathbf N^*,$$且(x_1=3)，求数列({x_n})的通项公式．
解答:
根据题意，有$$x_{n+1}=dfrac{(n+1)(n+2)}{n2}x_n+n+1,$$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$$dfrac{x_{n+1}}{(n+1)2(n+2)}+dfrac{1}{n+2}=dfrac{x_n}{n2(n+1)}+dfrac{1}{n+1},$$ 因此$$dfrac{x_n}{n2(n+1)}+dfrac{1}{n+1}=dfrac{x_{n-1}}{(n-1)2cdot n}+dfrac{1}{n}=cdots =dfrac{x_1}{2}+dfrac 12=2,$$所以(x_n=n^2(2n+1),ninmathbf 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.