每日一题[87] 构造辅助数列

数列$\left\{a_n\right\}$满足$a_1=1$，$a_{n+1}a_n+(n+2)a_{n+1}+na_n+n^2+2n+2=0$，求数列$\left\{a_n\right\}$的通项公式．

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解    根据已知，有

$$\left(a_{n+1}+n\right)\left(a_n+n+2\right)=-2,$$ 令$b_n=a_n+n+1$，则 $$\left(b_{n+1}-2\right)\left(b_n+1\right)=-2,$$ 整理得 $$\dfrac{2}{b_{n+1}}=\dfrac 1{b_n}+1,$$ 即 $$\dfrac{2^{n+1}}{b_{n+1}}=\dfrac{2^n}{b_n}+2^n,$$ 令$c_n=\dfrac{2^n}{b_n}$，得 $$c_{n+1}=c_n+2^n,$$ 而 $$c_1=\dfrac 2{b_1}=\dfrac 2{a_1+2}=\dfrac 23,$$ 从而 $$c_n=\dfrac 23+2^n-2=2^n-\dfrac 43,$$ 因此 $$b_n=\dfrac{2^n}{2^n-\dfrac 43},$$ 进而可得 $$a_n=\dfrac 4{3\cdot 2^n-4}-n.$$