math-ai/BlueMO

1{2    "source_file": "./raw_volume-zh/volume11/exercise3.tex",3    "problem_type": "calculation",4    "problem": "问题1. 试用母函数方法求下列数列的通项 $a_n$.\n(1) $a_0=2, a_1=5, a_{n+2}=3 a_{n+1}-2 a_n(n=0,1,2, \\cdots)$;\n(2) $a_1=3, a_2=15, a_{n+2}=a_{n+1}+6 a_n-12(n=1,2, \\cdots)$.",5    "solution": "(1) 设 $f(x)=a_0+a_1 x+a_2 x^2+\\cdots+a_n x^n+\\cdots$, 则\n$$\n\\begin{array}{rr}\n-3 x f(x)= & -3 a_0 x-3 a_1 x^2-\\cdots-3 a_{n-1} x^n-\\cdots, \\\\\n2 x^2 f(x)= & 2 a_0 x^2+\\cdots+2 a_{n-1} x^n+\\cdots .\n\\end{array}\n$$\n三式相加, 并利用 $a_0=2, a_1=5, a_n-3 a_{n-1}+2 a_{n-2}=0(n=2,3, \\cdots)$\n得 $\\left(1-3 x+2 x^2\\right) f(x)=2-x$, 所以 $f(x)=\\frac{2-x}{1-3 x+2 x^2}= \\frac{3(1-x)-(1-2 x)}{(1-x)(1-2 x)}=\\frac{3}{1-2 x}-\\frac{1}{1-x}=3 \\sum_{n=0}^{\\infty}(2 x)^n-\\sum_{n=0}^{\\infty} x^n=\\sum_{n=0}^{\\infty}\\left(3 \\cdot 2^n\\right. -1) x^n$, 所以 $a_n=3 \\cdot 2^n-1$.\n(2) 设 $f(x)=a_0+a_1 x+a_2 x^2+\\cdots+a_n x^n+\\cdots$, 则\n$$\n\\begin{array}{rlrl}\n-x f(x) & = & -a_0 x-a_1 x^2-\\cdots-a_{n-1} x^n-\\cdots, \\\\\n-6 x^2 f(x) & = & -6 a_0 x^2-\\cdots-6 a_{n-2} x^n-\\cdots, \\\\\n\\frac{12}{1-x} & =12+12 x+12 x^2+\\cdots+12 x^n+\\cdots .\n\\end{array}\n$$\n四式相加, 并利用 $a_0=\\frac{1}{6}\\left(a_2-a_1+12\\right)=4, a_1=3, a_n-a_{n-1}-6 a_{n-2}+ 12=0(n=2,3, \\cdots)$ 得 $\\left(1-x-6 x^2\\right) f(x)+\\frac{12}{1-x}=16+11 x, f(x)= \\frac{4-5 x-11 x^2}{\\left(1-x-6 x^2\\right)(1-x)}=\\frac{4-5 x-11 x^2}{(1-x)(1+2 x)(1-3 x)}=\\frac{A}{1-x}+\\frac{B}{1+2 x}+ \\frac{C}{1-3 x}$, 于是 $A=\\left.\\frac{4-5 x-11 x^2}{(1+2 x)(1-3 x)}\\right|_{x=1}=2, B=\\left.\\frac{4-5 x-11 x^2}{(1-x)(1-3 x)}\\right|_{x=-\\frac{1}{2}} =1, C=\\left.\\frac{4-5 x-11 x^2}{(1-x)(1+2 x)}\\right|_{x=\\frac{1}{3}}=1$, 所以 $f(x)=\\frac{2}{1-x}+\\frac{1}{1+2 x}+\\frac{1}{1-3 x}$. $=2 \\sum_{n=0}^{\\infty} x^n+\\sum_{n=0}^{\\infty}(-2 x)^n+\\sum_{n=0}^{\\infty}(3 x)^n=\\sum_{n=0}^{\\infty}\\left(3^n+(-2)^n+2\\right) x^n$, 故得 $a_n=3^n +(-2)^n+2$.",6    "remark": "",7    "figures": []8}