Question

已知an+3=((an+1*an+2)+1)\an,求通项

Answer 1

设bn=(an+1)*an
则(an+3)*an=((an+1)*(an+2))+1
(an+3)/(an+1) =(an+2)/(an)+1/((an+1)*(an))
(bn+2)/(bn+1)=(bn+1)/(bn)+1/bn
(bn+2)*(bn)=((bn+1)+1)*(bn+1)
1/(bn+1)-1/((bn+1)+1)=1/((bn+2)*(bn))
(bn+2)/(bn+1)-(bn)/(bn+1)-(bn+2)/((bn+1)+1)+(bn)/((bn+1)+1)=1/(bn)-1/(bn+2)
(bn+2)/(bn+1)+(bn)/((bn+1)+1)+1/(bn+2)=(bn)/(bn+1)+(bn+2)/((bn+1)+1)+1/(bn)
注意到(bn+2)*(bn)/(((bn+1)+1)*(bn+1))=1
故(bn+2)/(bn+1)+(bn)/(bn+2)+1/(bn+2)=(bn)/(bn+1)+(bn+1)/(bn)+1/(bn)
(bn+2)/(bn+1)+(bn)/(bn+2)+1/(bn+2)+1/(bn+1)=(bn)/(bn+1)+(bn+1)/(bn)+1/(bn)+1/(bn+1)
故对任意的n有(bn)/(bn+1)+(bn+1)/(bn)+1/(bn)+1/(bn+1)=a (a为常数)
(bn)^2+(bn+1)^2+(bn)+(bn+1)=a(bn)*(bn+1)
(bn+1)^2+(bn+2)^2+(bn+1)+(bn+2)=a(bn+1)*(bn+2)
则 (bn+2),(bn)为一二次方程两解,由伟达定理 (bn+2)+(bn)=a* (bn+1)-1
(bn+2)+(((a^2-4)^0.5-a)/2)*(bn+1)+2/(2-(a^2-4)^0.5-a)=(((a^2-4)^0.5-a)/2)*((bn+1)+(((a^2-4)^0.5-a)/2)*(bn)+2/(2-(a^2-4)^0.5-a))
故对任意的n有(bn+1)+(((a^2-4)^0.5-a)/2)*(bn)+2/(2-(a^2-4)^0.5-a)=m*(((a^2-4)^0.5-a)/2)^n (m为常数)
接下来就可以求出bn,进而求出an了.