Question

1/2+1/4+1/8+1/16+1/32+1/64+1/125+1/256+1/512

Answer 1

等比数列,有求和公式
如果没学过,可以用错位相减
令m=1/2+1/4+1/8+...+1/512
同时乘2,得:
2m=1+1/2+1/4+...+1/256
相减,得:
2m-m=(1+1/2+1/4+...+1/256)-(1/2+1/4+...+1/512)
m=1-1/512=511/512

追问

请问等比数列的公式

追答

Sn=a1(1-q^n)/(1-q) =(a1-an×q)/(1-q) (q≠1)

Answer 2

这个是等比数列的加法。求和公式：Sn=n*a1(q=1)
Sn=a1(1-q^n)/(1-q)
=(a1-a1q^n)/(1-q)
=a1/(1-q)-a1/(1-q)*q^n ( 即a-aq^n)
(前提：q不等于 1) 所以结果是1-(1/2)^9=511/512

Answer 3

解：这是首项为1/2，公比为1/2的等比数列，
一共有9项，
则原式=1/2[1-(1/2)^9]/(1-1/2)
=1-(1/2)^9
=1-1/512
=511/512

Answer 4

1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512
=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/512-1/512
=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/256-1/512
=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/128-1/512
=1/2+1/4+1/8+1/16+1/32+1/64+1/64-1/512
=...
=1/2+1/2-1/512
=1-1/512
=511/512

Answer 5

1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512
=（1-1/2）+（1/2-1/4）+（1/4-1/8）+（1/8-1/16）+（1/16-1/32）+（1/32-1/64）+（1/64-1/128）+（1/128-1/256）+（1/256-1/512）
=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128+1/128-1/256+1/256-1/512
=1-1/512
=511/512