计算1⼀2+1⼀4+1⼀8+1⼀16+1⼀32+1⼀64+1⼀128怎么简便计算

2025-01-03 07:36:32
推荐回答(5个)
回答1:

原式: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)
=1-1/2+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……互相抵消
=127/128
(望采纳,谢谢)

回答2:

原式:1/2+1/4+1/8+1/16+1/32+1/64+1/128
=1-1/128
=127/128
(望采纳,谢谢)

回答3:

原式=1/2+1/4+……+1/128+1/128-1/128 (1/128/128=1/64)
=1/2+1/4+……+1/32+1/64+164-1/128(1/64+164=1/32)
=……(抵消下去)
=1/2+1/2-1/128
= 1-1/128
=127/128

希望对你有帮助 ,我尽量写得更加详细。

回答4:

等比数列前n项和公式

1/2+1/4+1/8+1/16+1/32+1/64+1/128
=1/2+(1/2)^2+(1/2)^3+……+(1/2)^7
=1/2×[1-(1/2)^7]/(1-1/2)
=1-(1/2)^7
=1-1/128
=127/128

回答5:

通分啊 ,分母用128