设数列{an}的前n项和为Sn,且a1=2,Sn=nan+n(n+1),1.求an 2.设Tn为数列an/2^n的前N项和,求Tn
问题描述：
设数列{an}的前n项和为Sn,且a1=2,Sn=nan+n(n+1),1.求an 2.设Tn为数列an/2^n的前N项和,求Tn
答
\Sn=nan+n(n+1)
S(n-1)=(n-1)a(n-1)+n(n-1)
an=Sn-S(n-1)=nan-(n-1)a(n-1)+2n
(n-1)an=(n-1)a(n-1)-2n
an=a(n-1)-2n/(n-1)
an-a(n-1)=-2/n-1
......
a3-a2=-2x3/3-1
a2-a1=-2x2/2-1
an-a1=-2(n/n-1+n-1/n-2+.....+3/3-1+2/n-1)
an-a1=-2[n(n+1)/2x(1/n-2-1/n-1+......1/2-1/3+1-1/2)]
an=-(n+1)n(n-2)/n-1
bn=-n(n+1)(n-2)x2^n/n-1
Tn=
后面的自己慢慢算吧