数列.(1720:16:46)设2次方程anx^2-an+1x+1=0(n=1,2,3,……)有两根b和c,且满足6b-2bc+6c=3 ({an} {an+1} 是数列 )试用an表示an+1求证{an-2/3}是等比数列

Question

数列.(17 20:16:46)
设2次方程anx^2-a n+1 x+1=0 (n=1,2,3,……)有两根 b 和c,且满足6b-2bc+6c=3  ({an}  {an+1}  是数列 )
试用an表示an+1
求证{an- 2/3}是等比数列

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...