Question

已知Sn是数列{an}的前n项和,且a1=1,nan+1=2sn(n属于N*),设数列{bn}满足

已知Sn是数列{an}的前n项和,且a1=1,nan+1=2sn(n属于N*),设数列{bn}满足bn=2/(n+2)an,求数列{bn}的前n项和

Answer 1

na(n+1)=2sn
(n-1)an=2s(n-1) 两式相减得
na(n+1)-(n-1)an=2(sn-s(n-1)=2an
na(n+1)=(n-1)an+2an=(n+1)an
a(n+1)/an=(n+1)/n
an/a(n-1)=n/(n-1)
....................
a2/a1=2/1
左式对应相乘，右式对应相乘得
a(n+1)/a1=(n+1)/1
a(n+1)=(n+1)
an=n

bn=2/(n+2)an
=2/n(n+2)
=1/n -1/(n+2)
Tn=b1+b2+..........+bn
=1-1/3+1/3-1/5+..........+1/n-1/(n+2)
=1-1/(n+2)
=(n+1)/(n+2)

Answer 2