p 已知数列{an}满足a1=1/2,an+1=（n+1/2n）an,记数列{an}的前n项和为sn.求数列an通项公式

a(n+1)=[(n+1)/(2n)]an
a(n+1)/(n+1)=(1/2)(an/n)
[a(n+1)/(n+1)]/(an/n)=1/2,为定值
a1/1=1/2/1=1/2,数列{an/n}是以1/2为首项,1/2为公比的等比数列
an/n=(1/2)(1/2)^(n-1)=1/2ⁿ
an=n/2ⁿ
数列{an}的通项公式为an=n/2ⁿ
Sn=a1+a2+a3+...+an=1/2+2/2²+3/2³+...+n/2ⁿ
Sn/2=1/2²+2/2³+...+(n-1)/2ⁿ+n/2^(n+1)
Sn-Sn/2=Sn/2=1/2+1/2²+...+1/2ⁿ -n/2^(n+1)
Sn=1+1/2+...+1/2^(n-1) -n/2ⁿ
=1×(1-1/2ⁿ)/(1-1/2) -n/2ⁿ
=2 -(n+2)/2ⁿ