生成函数
使用母函数的方法求谢列数列的通项 \(a_n.\)
\((1) a_0=2,a_1=5,a_{n+2}=3a_{n+1}-2a_n(n=0,1,2,\cdots);\)
解: 设 \(f(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots.\) 则:
\(\qquad-3f(x)=-3a_0x-3a_1x^2-3a_2x^3-\cdots.\)
\(\quad\quad\qquad\qquad2f(x)=+2a_0x^2+2a_1x^3+2a_2x^4-\cdots.\)
三式相加并结合 \(a_0=2,a_1=5,a_{n+2}=3a_{n+1}-2a_n\) 可知:
\(f(x)=(1-3x+2x^2)f(x)=2-x \qquad \qquad \qquad \qquad (*)\)
解出来 \(f(x)=\dfrac{2-x}{1-3x+2x^2}=\dfrac{2-x}{(x-1)(2x-1)}\)
设 \(f(x)=\dfrac{A}{x-1}+\dfrac{B}{2x-1}=\dfrac{2-x}{(x-1)(2x-1)}\)
\((*)\) 式两边乘 \(x-1\) 后,令 \(x=1\) 得:
\(A=\dfrac{2-x}{2x-1}\Big{|}_{x=1}=1\)
\((*)\) 式两边乘 \(2x-1\) 后,令 \(x=\dfrac{1}{2}\) 得:
\(B=\dfrac{2-x}{x-1}\Big{|}_{x=\dfrac{1}{2}}=-3\)
\[f(x)=-\dfrac{1}{x-1}-\dfrac{3}{2x-1}=\sum_{n=1}^\infty\left[3\times2^n-1\right]x^n \]\(\therefore a_n = 3\times 2^n -1\)
标签:组合,dfrac,2x,cdots,数学,2a,qquad,3a From: https://www.cnblogs.com/chihirofujisaki/p/18125108