数列1,2+1/2,3+1/2+1/4,……,n+1/2+1/4+……+1/(2的n-1次方)的前n项和为?
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解决时间 2021-03-22 22:09
- 提问者网友:无心恋土
- 2021-03-22 02:30
数列1,2+1/2,3+1/2+1/4,……,n+1/2+1/4+……+1/(2的n-1次方)的前n项和为?
最佳答案
- 二级知识专家网友:修女的自白
- 2021-03-22 03:38
n+1/2+1/4+……+1/(2的n-1次方)=n+1-1/2^(n-1)
sn=(1+1-1/2^0)+(2+1-2^1)+.....(n+1-1/2^(n-1))
=1+2+3+......n+(1+1+......1)(n个1)-1/2^0-1/2^1+。。。。。1/2^n-1
=n(n+1)/2+n-(2-1/2^n-1)
=n(n+3)/2+1/2^(n-1)-2
注,1/2+1/4+……+1/(2的n-1次方)=1-1/(2的n-1次方)
不信你1,2,3代进去试
sn=(1+1-1/2^0)+(2+1-2^1)+.....(n+1-1/2^(n-1))
=1+2+3+......n+(1+1+......1)(n个1)-1/2^0-1/2^1+。。。。。1/2^n-1
=n(n+1)/2+n-(2-1/2^n-1)
=n(n+3)/2+1/2^(n-1)-2
注,1/2+1/4+……+1/(2的n-1次方)=1-1/(2的n-1次方)
不信你1,2,3代进去试
全部回答
- 1楼网友:情战辞言
- 2021-03-22 04:52
1+2+1/2+3+1/2+1/4+……+n+1/2+1/4+……+n+1/(2^(n-1)
=(1+2+3+……+n)+(1/2+1/4+……+1/2^(n-1)
=1/2*n*(n+1)+1/2*[1-(1/2)^(n-1-1)]/(1-1/2)
=1/2*n*(n+1)+1-4/2^n
- 2楼网友:浪者不回头
- 2021-03-22 03:53
1,2+1/2,3+1/2+1/4,…,n+1/2+1/4+…+1/2n-1 最后一项应该是1/2^(n-1)吧。
sn=1+2+1/2+3+1/2+1/4+…+n+1/2+1/4+…1/2^(n-1)
=1+2+3+...+n+1/2+1/2+1/4+…+n+1/2+1/4+…1/2^(n-1)
=n(n+1)/2+(1-1/2)+(1-1/4)+...+[1-1/2^(n-1)]
=n(n+1)/2+1-1/2+1-1/4+...+1-1/2^(n-1)
=n(n+1)/2+(n-1)-[1/2+1/4+...+1/2^(n-1)]
=n(n+1)/2+(n-1)-[1-1/2^(n-1)]
=n(n+1)/2+n-1-1+1/2^(n-1)
=n(n+1)/2+n+1/2^(n-1)
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