数列an中,Sn=n*（2n-1）*an 求其通项an和前n项和Sn

Sn=n*（2n-1）*an
S(n-1)=(n-1)[2(n-1)-1]*a(n-1)
相减,Sn-S(n-1)=an
an=n*（2n-1）*an-(n-1)(2n-3)a(n-1)
[n(2n-1)-1]an=(n-1)(2n-3)a(n-1)
(2n+1)(n-1)an=(n-1)(2n-3)a(n-1)
an/a(n-1)=(2n-3)/(2n+1)
所以a(n-1)/a(n-2)=[2(n-1)-3]/[2(n-1)+1]=(2n-5)/(2n-1)
a(n-2)/a(n-3)=(2n-7)/(2n-3)
……
a4/a3=5/9
a3/a2=3/7
a2/a1=1/5
相乘
an/a1=3*1/(2n+1)(2n-1)
an=9/(4n^2-1)
an=9/(2n+1)(2n-1)=(9/2)[1/(2n-1)-1/(2n+1)]
所以Sn=(9/2)*[1/1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-1)-1/(2n+1)]
=(9/2)[1-1/(2n+1)]
=9n/(2n+1)