![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
求S=1/2^2+2/2^3+3/2^4+4/2^5+...+n/2^(n+1)的值的解题过程
2个回答
展开全部
求S=1/2^2+2/2^3+3/2^4+4/2^5+...+n/2^(n+1)的值的解题过程
解:S=1/2^2+2/2^3+3/2^4+4/2^5+...+n/2^(n+1),两边同乘以1/2
(1/2)S=1/2^3+2/2^4+3/2^5+4/2^6+...+n/2^(n+2),两式相减
S-(S/2)=S=1/2^2+1/2^3+1/2^4+1/2^5+...+1/2^(n+1)-n/2^(n+2),
S/2=[1/2^2+1/2^3+1/2^4+1/2^5+...+1/2^(n+1)]-n/2^(n+2)
S/2=[1/2^2(1-1/2^n)/(1-1/2)]-n/2^(n+2)
S/2=[1/2^2(1-1/2^n)/(1/2)]-n/2^(n+2)
S/2=[(1/2)(1-1/2^n)]-n/2^(n+2)
S=(1-1/2^n)-n/2^(n+2)
S=1-1/2^n-n/2^(n+2)
解:S=1/2^2+2/2^3+3/2^4+4/2^5+...+n/2^(n+1),两边同乘以1/2
(1/2)S=1/2^3+2/2^4+3/2^5+4/2^6+...+n/2^(n+2),两式相减
S-(S/2)=S=1/2^2+1/2^3+1/2^4+1/2^5+...+1/2^(n+1)-n/2^(n+2),
S/2=[1/2^2+1/2^3+1/2^4+1/2^5+...+1/2^(n+1)]-n/2^(n+2)
S/2=[1/2^2(1-1/2^n)/(1-1/2)]-n/2^(n+2)
S/2=[1/2^2(1-1/2^n)/(1/2)]-n/2^(n+2)
S/2=[(1/2)(1-1/2^n)]-n/2^(n+2)
S=(1-1/2^n)-n/2^(n+2)
S=1-1/2^n-n/2^(n+2)
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
S=1/2^2+2/2^3+3/2^4+4/2^5+...+n/2^(n+1)
则2S=1/2+2/2^2+3/2^3+4/2^4+...+n/2^n
2式相减得:
S=2S-S
=1/2+1/2^2+2/2^3+3/2^4+...+1/2^n-n/2^(n+1)
=1/2*[1-1/2^(n+1)]/(1-1/2)-n/2^(n+1)
=4-1/2^(n-1)-n/2^(n+1)
则2S=1/2+2/2^2+3/2^3+4/2^4+...+n/2^n
2式相减得:
S=2S-S
=1/2+1/2^2+2/2^3+3/2^4+...+1/2^n-n/2^(n+1)
=1/2*[1-1/2^(n+1)]/(1-1/2)-n/2^(n+1)
=4-1/2^(n-1)-n/2^(n+1)
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