Question

1/2+1/4+1/8+1/16+1/32+1/64+1/128[简便]？

Answer 1

方法1、

1/2+1/4+1/8+1/16+1/32+1/64+1/128

=（1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/128）-1/128.............倒过来计算

=1-1/128

=127/128

方法2、

1/2+1/4+1/8+1/16+1/32+1/64+1/128

=（1/2+1/4+1/8+1/16+1/32+1/64+1/128）×2

             -(1/2+1/4+1/8+1/16+1/32+1/64+1/128)

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

=1-1/128

=127/128

Answer 2

很显然再加上1/128的话
就不断得到1/128+1/128=1/64
1/64+1/64=1/32等等
最后得到1/2+1/2 -1/128
也就是原式=1 -1/128
所以式子的结果为127/128

Answer 3

方法1、
1/2+1/4+1/8+1/16+1/32+1/64+1/128
=（1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/128）-1/128.............倒过来计算
=1-1/128
=127/128
方法2、
1/2+1/4+1/8+1/16+1/32+1/64+1/128
=（1/2+1/4+1/8+1/16+1/32+1/64+1/128）×2
             -(1/2+1/4+1/8+1/16+1/32+1/64+1/128)
=(1+1/2+1/4+1/8+1/16+1/32+1/64)-(1/2+1/4+1/8+1/16+1/32+1/64+1/128)
=1-1/128
=127/128

Answer 4

1/2+1/4+1/8+1/16+1/32+1/64+1/128的简便算法。
这是一个等比数列，直接用等比数列的求和公式就可以了。
Sn=a1(1-q^n)/(1-q)=1-1/128=127/128

Answer 5

1/2+1/4+1/8+1/16+1/32+1/64+1/128
=(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)
=1-1/128
=127/128
望采纳噢~