Question

数列.(17 20:16:46)

Answer 1

anx^2-a(n+1)x+1=0b+c=a(n+1)/anbc=1/an6b-2bc+6c=6(b+c)-2bc=6[a(n+1)/an]-2/an=36a(n+1)-2=3ana(n+1)=(1/2)an+(1/3) a(n+1)=(1/2)an+(1/3)a(n+1)-2/3=(1/2)an+(1/3)-2/3=(1/2)an-1/3=(1/2)[an-2/3][a(n+1)-2/3]/[a...