Question

数列｛an｝的前n项和为sn,且满足a1=1,2s=(n+1)an,求数列｛an｝的通项,求和wn=1/（a2^2-1）

Answer 1

2·a(n) = 2[Sn - S(n-1)] = (n+1)an - n·a(n-1)

∴(n-1)an = n·a(n-1) ,

∴an/[a(n-1)] = n/(n-1) ,. ,a3/a2 = 3/2 ,a2/a1 = 2/1 ,将上述式子相乘：an/a1 = n ,∴an = n·1 = n ,

即：{an}是以1为首项、1为公差的等差数列 .

2)

(n+1)^2-1=(n+2)*n,

得wn=1/(1*3)+.1/(n+2)*n=1/2*(1-1/3+1/2-1/4+1/3-1/5.1/n-1/(n+2))

=3/4-1/2[1/(n+1)+1/(n+2) ]