MT129常数变易法

Posted 2020-10-30 青春的记忆

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．

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.