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

来源：百度知道 编辑：UC知道 时间：2024/05/10 05:42:56
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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

解：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……