简单推式子...
1
已知 \(a_n = 2 \times a_{n - 1} + 3 \times a_{n - 2} + 3^{n}(n \ge 2)\) , \(a_0 = -1\) , \(a_1 = 1\) , 求 \(a_n\) .
解：
设 \(f(x) = \sum\limits_{i = 0}a_ix^i\)
则
\[\begin{alignedat}{3} f(x) & = \sum\limits_{i = 0}a_ix^i \\ 2x \times f(x) & = \sum_{i = 1} 2a_{i - 1}x^{i} \\ 3x^2 \times f(x) & = \sum_{i = 2} 3a_{i - 2}x^{i} \end{alignedat}\]三式相减，得：
\[\begin{alignedat}{3} (1 - 2x - 3x^2)f(x) & = -1 + \sum\limits_{i = 1}3^ix^i \\ & = -1 + \frac{1}{1 - 3x} - 1 \\ & = \frac{6x - 1}{1 - 3x} \end{alignedat}\]所以可知
\[f(x) = \frac{6x - 1}{(1 + x)(1 - 3x)^{2}} \]将其转化成多个等比数列相加形式
\[f(x) = \frac{A}{1 + x} + \frac{B}{(1 - 3x)^2} + \frac{C}{1 - 3x} \]令 \(x = -1\) 等式两端同乘 \(x + 1\) , 得:
\[\frac{6x - 1}{(1 - 3x)^2} = A \]解得 \(A = -\frac{7}{16}\)
同理令 \(x = \frac{1}{3}\) , 解得 \(B = \frac{3}{4}\)
令 \(x \to +\infty\) , 并等式两端同乘 \(x\) , 得
\[\begin{alignedat}{3} & = \lim_{x \to +\infty} xf(x) \\ & = \lim_{x \to +\infty} \left(\frac{xA}{1 + x} + \frac{xB}{(1 - 3x)^2} + \frac{xC}{1 - 3x}\right) \end{alignedat}\]\(\because \lim\limits_{x \to +\infty}\frac{x}{1 + x} = 1\) , \(\lim\limits_{x \to +\infty}\frac{x}{1 - 3x} = -\frac{1}{3}\) ,
\(\lim\limits_{x \to +\infty}\frac{6x - 1}{(1 - 3x)^2} = 0\)
\(\therefore 0 = A - \frac{1}{3}C\) ,
\(C = - \frac{21}{16}\)