1/2+1/4+1/8+1/16+1/32+……+1/1024+1/2048+1/2^n

1/2+1/4+1/8+1/16+1/32+……+1/1024+1/2048+1/2^n
1/2+1/4+1/8+1/16+1/32+……+1/1024+1/2048+1/2^n

1/2+1/4+1/8+1/16+1/32+……+1/1024+1/2048+1/2^n
等比数列,有求和公式 Sn=a1(1-q^n)/(1-q) =(a1-an×q)/(1-q) (q≠1)
如果没学过,可以用错位相减
令m=1/2+1/4+1/8+...+1/2^n
同时乘2,得:
2m=1+1/2+1/4+...+1/2^（n-1）
相减,得:
2m-m=(1+1/2+1/4+...+1/2^n)-(1/2+1/4+...+1/2^（n-1）)
m=1-1/2^n
学习宝典竭诚为您解答,希望能够获得采纳.

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