求和Sn=1+(1+1/2)+(1+1/2+1/4)+.+[1+1/2+1/4.+1/2^(n-1)]
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![求和Sn=1+(1+1/2)+(1+1/2+1/4)+.+[1+1/2+1/4.+1/2^(n-1)]](/uploads/image/z/2181378-66-8.jpg?t=%E6%B1%82%E5%92%8CSn%3D1%2B%281%2B1%2F2%29%2B%281%2B1%2F2%2B1%2F4%29%2B.%2B%5B1%2B1%2F2%2B1%2F4.%2B1%2F2%5E%28n-1%29%5D)
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……