{an}满足a1=3，an-2anan+1-an+1=0，求an。
热心网友
{an}满足a1=3，an-2ana(n+1)-a(n+1)=0，求an。an-2an*a(n+1)-a(n+1)=01/a(n+1)-1/an=2∴1/an是以1/3为首项、2位公差的等差数列1/an=1/3+2(n-1)an=1/[1/3+2n-2]=3/(6n-5)
热心网友
∵an-2anan+1-an=0∴1/ an+1-1/ an=2令bn+1=1/ an+1∴bn+1- bn =2∴bn是以b1为首项、2位公差的等差数列∵a1=3,∴b1=1/3,bn =1/3+2(n-1)=2n-5/3=(6n-5)/3∴an=1/bn=3/(6n-5)
热心网友
an-2anan+1-an+1=0a1-2a1a2-a2=03-2*3*a2-a2=03-6a2-a2=03=7a2a2=3/7a2-2a2a3-a3=03/7-2*(3/7)*a3-a3=03/7-(13/7)*a3a3=3/13same,a4=3/19.......from 3/1,3/7,3/13,3/19.......get 1,7,13,19.......1+6(n-1).....an=3/[1+6(n-1)]