求和Sn=1+(1+1/2)+(1+1/2+1/4)+.+[1+1/2+1/4.+1/2^(n-1)]

来源：学生作业帮助网编辑：六六作业网时间：2024/11/14 10:54:23

求和Sn=1+(1+1/2)+(1+1/2+1/4)+.+[1+1/2+1/4.+1/2^(n-1)]求和Sn=1+(1+1/2)+(1+1/2+1/4)+.+[1+1/2+1/4.+1/2^(n-1

求和Sn=1+(1+1/2)+(1+1/2+1/4)+.+[1+1/2+1/4.+1/2^(n-1)]
求和Sn=1+(1+1/2)+(1+1/2+1/4)+.+[1+1/2+1/4.+1/2^(n-1)]

求和Sn=1+(1+1/2)+(1+1/2+1/4)+.+[1+1/2+1/4.+1/2^(n-1)]
Sn=(2-1)+(2-1/2)+(2-1/4)+.{2-1/2^(n-1)}
=2*n-[1+1/2+1/4+.1/2^(n-1)]
=2n-{2-1/2^(n-1)]
=2n-2+1/2^(n-1)
要加油哦

1+1/2+1/4+.....+1/2^(n-1)
=1+1/2+1/4+.....+1/2^(n-2)+1/2^(n-1)+1/2^(n-1)-1/2^(n-1)
=1+1/2+1/4+.....+1/2^(n-2)+1/2^(n-2)-1/2^(n-1)
=.....
=2-1/2^(n-1)
故
Sn=（2-1）+（2-1/2）+（2-1/4）+……+[2-1/2^(n-1)]
=2n-[1+1/2+1/4+.....+1/2^(n-1)]
=2n-[2-1/2^(n-1)]
=2n-2+1/2^(n-1)

[1+1/2+1/4.....+1/2^(n-1)]=(1-1/2^n)/(1-1/2)(等比数列求和公式）
所以有：原式=1+2*(1-1/2^2)+2*(1-1/2^3)+2*(1-1/2^4)+……+2*(1-1/2^n)=1+(2+2+2+2+……+2)-[1/2+1/2^2+1/2^3+1/2^4+……+1/2^(n-1))]=1+2*(n-1)-[1-1/2^(n-1)]=2*(n-1)+1/2^(n-1)
头晕中……

求和Sn=1-2 3-4+ 数列Bn=1/n,求和Sn 数列求和:sn=1+1/2+1/3+…+1/n,求sn Sn求和 Sn=1+2x3+3x9+4x27+...+nx3的n-1次方高中数列求和An=1/n,求Sn. 无穷级数证明题正项级数an Sn= 求和an 发散求证求和 (a(n+1)/Sn) 也发散数列求和 bn=4/(n+1)(n+2)数列求和bn=4/(n+1)(n+2),Sn=? 数列求和用分组求和及并项法求和 Sn=1^2-2^2+3^2-4^2+…+(-1)^(n-1)·n^2 求和,Sn=1+11+111+1111+……+（n个1）求Sn 数列求和：Sn=1/1*2*3+1/2*3*4+.+1/n*(n+1)*(n+2) 求Sn 利用等差数列求和公式Sn=n(a1+an)/2证明Sn=na1+n(n-1)/2*d 求和：sn=1/2^2-1+1/4^2-1+.1/(2n)^2-1 求和Sn=1/1*4+1/4*7+.1/(3n-2)(3n+1) 求和Sn=1/(1*4)+1/(2*7)+.+1/n*(3n+1) 求和Sn=1/(1*4)+1/(2*7)+.+1/n*(3n+1) 求和Sn=1-3+5-7+...+(-1)^(n-1)(2n-1) sn=1*n+2(n-1)+3(n-2)+……+n*1 求和求和：Sn=1*n+2*(n-1)+3*(n-2)+……+n*1