a n ​ −4a n−1 ​ +4a n−2 ​ =n 2 \displaystyle\sum_{n=2}^{\infin}a_nx^n-4\displaystyle\sum_{n=2}^{\infin}a_{n-1}x^{n}+4\displaystyle\sum_{n=2}^{\infin}a_{n-2}x^{n}=\displaystyle\sum_{n=2}^nn^2x^n n=2 ∑ ∞ ​ a n ​ x n −4 n=2 ∑ ∞ ​ a n−1 ​ x n +4 n=2 ∑ ∞ ​ a n−2 ​ x n = n=2 ∑ n ​ n 2 x n From the table \displaystyle\sum_{n=0}^nn^2x^n=0+x+2^2x^2+3^2x^3+...=\dfrac{x(x+1)}{(1-x)^3} n=0 ∑ n ​ n 2 x n =0+x+2 2 x 2 +3 2 x 3 +...= (1−x) 3 x(x+1) ​ G(x)-2-5x-4x(G(x)-2)+4x^2G(x)G(x)−2−5x−4x(G(x)−2)+4x 2 G(x) =\dfrac{x(x+1)}{(1-x)^3}-x= (1−x) 3 x(x+1) ​ −x G(x)(1-4x+4x^2)=\dfrac{x(x+1)}{(1-x)^3}-4x+2G(x)(1−4x+4x 2 )= (1−x) 3 x(x+1) ​ −4x+2 G(x)=\dfrac{x^2+x}{(1-x)^3(1-2x)^2}+\dfrac{2}{1-2x}G(x)= (1−x) 3 (1−2x) 2 x 2 +x ​ + 1−2x 2 ​ \dfrac{x^2+x}{(1-x)^3(1-2x)^2}=\dfrac{A}{(1-x)^3}+\dfrac{B}{(1-x)^2}+\dfrac{C}{1-x} (1−x) 3 (1−2x) 2 x 2 +x ​ = (1−x) 3 A ​ + (1−x) 2 B ​ + 1−x C ​ +\dfrac{D}{(1-2x)^2}+\dfrac{E}{1-2x}+ (1−2x) 2 D ​ + 1−2x E ​ A(1-2x)^2+B(1-x)(1-2x)^2A(1−2x) 2 +B(1−x)(1−2x) 2 +C(1-x)^2(1-2x)^2+D(1-x)^3+C(1−x) 2 (1−2x) 2 +D(1−x) 3 +E(1-x)^3(1-2x)=\dfrac{x^2+x}{(1-x)^3(1-2x)^2}+E(1−x) 3 (1−2x)= (1−x) 3 (1−2x) 2 x 2 +x ​ x=0:A+B+C+D+E=0x=0:A+B+C+D+E=0 x=-1:9A+18B+36C+8D+24E=0x=−1:9A+18B+36C+8D+24E=0 x=1:A=2x=1:A=2 x=1/2:D=6x=1/2:D=6 x=2:9A-9B+9C-D+3E=6x=2:9A−9B+9C−D+3E=6 A=2, B=-3, C=-3, D=6, E=-2A=2,B=−3,C=−3,D=6,E=−2 G(x)=\dfrac{2}{(1-x)^3}-\dfrac{3}{(1-x)^2}-\dfrac{3}{1-x}G(x)= (1−x) 3 2 ​ − (1−x) 2 3 ​ − 1−x 3 ​ +\dfrac{6}{(1-2x)^2}-\dfrac{2}{1-2x}+\dfrac{2}{1-2x}+ (1−2x) 2 6 ​ − 1−2x 2 ​ + 1−2x 2 ​ \def\arraystretch{1.5} \begin{array}{c:c} \dfrac{2}{(1-x)^3} & 2\dbinom{n+2}{2} \\ \\ -\dfrac{3}{(1-x)^2} & -3(n+1)\\ \\ -\dfrac{3}{1-x} & -3 \\ \\ \dfrac{6}{(1-2x)^2} & 6(n+1)(2^n) \end{array} (1−x) 3 2 ​ − (1−x) 2 3 ​ − 1−x 3 ​ (1−2x) 2 6 ​ ​ 2( 2 n+2 ​ ) −3(n+1) −3 6(n+1)(2 n ) ​ a_n=2\dbinom{n+2}{2}-3(n+1)-3+6(2^n)(n+1)a n ​ =2( 2 n+2 ​ )−3(n+1)−3+6(2 n )(n+1)