Simone已知数列{an}的前n项和为Sn,且a1=1/2,a(n+1)=(n+1)an/2n,(1)求{an}的通项公式;(2)
的有关信息介绍如下：
解：(1)a(n+1)=(n+1)an/(2n)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)=1/2ⁿan=n/2ⁿ数列{an}的通项公式为an=n/2ⁿ(2)Sn=a1+a2+a3+...+an=1/2+2/2²+3/2³+...+n/2ⁿSn /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)=(1/2)(1-1/2ⁿ)/(1-1/2) -n/2^(n+1)=1- (n+2)/2^(n+1)Sn=2- (n+2)/2ⁿbn=n(2-Sn)=n[2-2+(n+2)/2ⁿ]=n(n+2)/2ⁿb1=1×3/2=3/2 b2=2×4/4=2 b3=3×5/8=15/8
