Question

a1=1,(n+1)an=2sn

1.求an的通项
2.设bn=an方-an-1方，求bn的前n项和为Tn

Answer 1

(n+1)an=2sn
na(n-1)=2s(n-1)
两式相减得
(n+1)an-na(n-1)=2an
(n+1)an-2an=na(n-1)
(n-1)an=na(n-1)

an/a(n-1)=n/(n-1)
an/a(n-1)=n/(n-1)
............
a3/a2=3/2
a2/a1=2/1
以上等式相乘得
an/a1=n
an=n

bn=(an)^2-[a(n-1)]^2
=n^2-(n-1)^2
=(n+n-1)(n-n+1)
=2n-1

Tn=b1+b2+.......+bn
=1+3+..........+2n-1
=(1+2n-1)*n/2
=n^2