1.
Aa(n+2)+Ba(n+1)+Can=0
設(shè)A[a(n+2)+xa(n+1)]=[a(n+1)+xan]
Aa(n+2)+(Ax-1)a(n+1)-xan]=0
B=Ax-1,C=-x
x=(B+1)/A=-C
(這里要求(B+1)=-AC)
設(shè)bn=a(n+1)+xan
bn是首項(xiàng)為a2+xa1,公比為1/A的等比數(shù)列,
bn=(a2+xa1)(1/A)^(n-1)
a(n+1)+xan=(a2+xa1)(1/A)^(n-1)
設(shè)[a(n+1)+y]+x[an+y]=0
a(n+1)+xan+y(x+1)=0
-y(x+1)=(a2+xa1)(1/A)^(n-1)
y=-[(a2+xa1)(1/A)^(n-1)]/(x+1)
所以an+y是以a1+y為首項(xiàng)-x為公比的等比數(shù)列,
an+y=(a1+y)(-x)^(n-1)
an=(a1+y)(-x)^(n-1)-y.
（將xy代入即可,方法如此,計(jì)算對(duì)否不能保證）
2.
a(n+2)=(Aan+B)/(Can+D),
設(shè)a(n+2)+x=y(an+x)/(Can+D),
a(n+2)=(yan+yx-xCan-xD)/(Can+D)
=[(y-xC)an+(yx-xD)]/(Can+D)
A=y-xC,B=yx-xD
x1={-A+D+√[(A-D)^2+4BC]}/(2C),y1=A+{-A+D+√[(A-D)^2+4BC]}/2
x2={-A+D-√[(A-D)^2+4BC]}/(2C),y2=A+{-A+D-√[(A-D)^2+4BC]}/2
a(n+2)+x1=y1(an+x1)/(Can+D)
a(n+2)+x2=y2(an+x2)/(Can+D)
兩式相除:
[a(n+2)+x1]/[a(n+2)+x2]=[y1(an+x1)/[y2(an+x2)]=(y1/y2)[(an+x1)/(an+x2)]
設(shè)bn=(an+x1)/(an+x2),則b1=(a1+x1)/(a1+x2),
b(n+2)=(y1/y2)bn
又設(shè)b(n+2)+Mb(n+1)=N[b(n+1)+Mbn]
整理得
b(n+2)+(M-N)b(n+1)=MNbn
M=N=±√(y1/y2)
b(n+1)+Mbn是公比為N=±√(y1/y2),首項(xiàng)為b2+Mb1的等比數(shù)列,
b(n+1)+Mbn=(b2+Mb1)[±√(y1/y2)]^(n-1)
再根據(jù)前面的方法求出(從“a(n+1)+xan=(a2+xa1)(1/A)^(n-1)”開始).
(關(guān)系式中是第n+2項(xiàng)與第n的關(guān)系,在已知條件中應(yīng)有a1和a2,否則求不出b2)