已知函数f(x)=(7x+5)/(x+1),数列an中,2an+1-2an+an+1an=0,a1=1,且an≠0,数列bn中,bn=f(an-1)⑴求证:数列{1/an}是等差数列⑵求数列{bn}的通项公式⑶求数列{|bn|}的前项和Sn

Question

已知函数f(x)=(7x+5)/(x+1),数列an中,2an+1-2an+an+1an=0,a1=1,且an≠0,数列bn中,bn=f(an-1)
⑴求证:数列{1/an}是等差数列
⑵求数列{bn}的通项公式
⑶求数列{|bn|}的前项和Sn

Answer 1

(1)
2a(n+1)-2an+a(n+1)an=0
1/a(n+1)-1/an= 1/2
{1/an}是等差数列, d=1/2
(2)
1/an -1/a1= (n-1)/2
an = 2/(n+1)
bn=f(an -1)
= f((1-n)/(1+n))
= (7[(1-n)/(1+n)]+5)/([(1-n)/(1+n)]+1)
= (12-2n)/2
=6-n
(3)
bn > 0
n< 6
for n>=6
|b1|+b2|+...+|bn| = (11-n)n/2
for n >7
|b1|+b2|+...+|bn|
=(b1+b2+...+b6) -(b7+b8+...+bn)
=15 + (n-7)(n-6)/2