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．
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\[a_n=\dfrac 4{3\cdot 2^n-4}-n.\]