数列Cn=n(1/2)^n,求前n项和Sn.
1个回答
展开全部
Sn=1*(1/2)+2*(1/2)^2+3*(1/2)^3+-----+(n-1)*(1/2)^(n-1)+n*(1/2)^n;
Sn/2=1*(1/2)^2+2*(1/2)^3+3*(1/2)^4+-----+(n-1)*(1/2)^n+n*(1/2)^(n+1);
Sn-Sn/2=1*(1/2)+(2-1)*(1/2)^2+(3-2)*(1/2)^3+-----+(n-n+1)*(1/2)^n-n*(1/2)^(n+1)
= Sn/2 =(1/2)+(1/2)^2+(1/2)^3+-----+(1/2)^n-n*(1/2)^(n+1)
=(1/2)(1-(1/2)^n)/(1-(1/2))-n*(1/2)^(n+1)=1-(1/2)^n-n*(1/2)^(n+1)
=1-(n+2)*(1/2)^(n+1)
Sn=2-(n+2)/2^n
Sn/2=1*(1/2)^2+2*(1/2)^3+3*(1/2)^4+-----+(n-1)*(1/2)^n+n*(1/2)^(n+1);
Sn-Sn/2=1*(1/2)+(2-1)*(1/2)^2+(3-2)*(1/2)^3+-----+(n-n+1)*(1/2)^n-n*(1/2)^(n+1)
= Sn/2 =(1/2)+(1/2)^2+(1/2)^3+-----+(1/2)^n-n*(1/2)^(n+1)
=(1/2)(1-(1/2)^n)/(1-(1/2))-n*(1/2)^(n+1)=1-(1/2)^n-n*(1/2)^(n+1)
=1-(n+2)*(1/2)^(n+1)
Sn=2-(n+2)/2^n
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询