Question

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?

Answer 1

n+1/2+.....1/2^(n-1)=n+1-1/2^(n-1)
so 前N项和为：
2-1+3-1/2+4-1/4.......+n+1-1/2^(n-1)
=(2+3+4......n+1)-(1+1/2+1/4+1/8.......1/2^(n-1))
=（n+2）(n+1)/2-3+1/2^(n-)1
附录：1+2+3......+n=n(n+1)/2
1+1/2+1/4.........+1/2^(n-1)=2-1/2^(n+1)
这是常识

Answer 2

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

Answer 3

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