極限fn(x)=X^n+2X^(n-1)被X^2-3X+2除

設Fn(x)=X^n+2X^(n-1)=(X^2-3X+2)Q(X)+An*X+Bn; 求：(1)Lim(n→∞)(An/Bn)=?Ans: Fn(x)=X^n+2X^(n-1)=(X-1)(X-2)Q(X)+(An*X+Bn)Fn(1)=1+2=3=0+An+Bn => An+Bn=3Fn(2)=2^n+2*2^(n-1)=2*2^n=2^(n+1)=0+2An+Bn => 2An+Bn=2^(n+1)兩者相減: An=2^(n+1)-3; Bn=6-2^(n+1)Lim(n→∞)(An/Bn)=[2^(n+1)-3]/[6-2^(n+1)]=-1........ans
(2)Σ(n=1~∞)An/(3^n)=?=[2^(n+1)-3]/3^n=2*(2/3)^n-3/3^n=2*(r+r^2+r^3+....+r^n)-3*(1/3+1/3^2+1/3^3+.....1/3^n)=2*r/(1-r)-3*(1/3)/[1-(1/3)]=(4/3)/(1/3)-(3/3)/(2/3)=4-3/2=5/2.........answhere r=2/3

f_n(x) 被 x^2-3x+2 除之後的餘式為 a_n x + b_n
因此, f_{n+1}(x) 被 x^2-3x+2 除之後的餘式, 同於 x(a_n x+b_n)
被除後的餘式, 即 (3a_n+b_n)x-2a_n. 換言之, 即
a_{n+1}x+b_{n+1} = (3a_n+b_n)x-2a_n
即: a_{n+1} = 3a_n+b_n, b_{n=1}=-2a_n.
又, a_1=1, b_1=2. 故 a_2 = 3*1+2 = 5.


a_{n+1} = 3a_n+b_n = 3a_n-2a_{n-1}, a_1=1,a_2=5
可解出 a_n=1+2^n, 故 b_n = -2a_{n-1} = -2(1+2^{n-1}) = -2-2^n

故
(1) a_n/b_n = (1+2^n)/(-2-2^n) → -1 當 n→∞
(2) Σa_n/3^n = Σ(1+2^n)/3^n = Σ(1/3)^n+Σ(2/3)^n
= (1/3)/(1-1/3) + (2/3)/(1-2/3) = 1/2+2 = 5/2.