1/2+1/4+1/8+1/16+1/32+1/64+1/128加省略号等于多少?
2个回答
展开全部
1/2+1/4+1/8+……+1/128
解:方法(一)
设原式为A
(1/2+1-4+1/8+1/16+……+1/128)×2为B
B一A得:
(1/2+1/4+1/8+1/16+……+1/128)×2-(1/2+1/4+1/8+1/16+……+1/128)
=1+1/2+1/4+1/8+1/16++……+1/64-1/2-1/4-1/8-1/16-1/32-……-1/128
=1+(1/2-1/2)+(1/4-1/4)+……+(1/64-1/64)-1/128
=1-1/128
=127/128
方法(二)
1/2+1/4+1/8+……+1/64+1/128
=1/2+1/4+1/8+……++1/64+1/128+1/128-1/28
=1/2+1/4+1/8+……1/64+1/64-1/128
=1/2+1/4+1/8+……+1/32+1/32-1/128
=1/2+1/4+1/8+1/16+1/16-1/128
=1/2+1/4+1/8+1/8-1/128
=1/2+1/4+1/4-1/128
=1/2+1/2-1/128
=1-1/128
=127/128
解:方法(一)
设原式为A
(1/2+1-4+1/8+1/16+……+1/128)×2为B
B一A得:
(1/2+1/4+1/8+1/16+……+1/128)×2-(1/2+1/4+1/8+1/16+……+1/128)
=1+1/2+1/4+1/8+1/16++……+1/64-1/2-1/4-1/8-1/16-1/32-……-1/128
=1+(1/2-1/2)+(1/4-1/4)+……+(1/64-1/64)-1/128
=1-1/128
=127/128
方法(二)
1/2+1/4+1/8+……+1/64+1/128
=1/2+1/4+1/8+……++1/64+1/128+1/128-1/28
=1/2+1/4+1/8+……1/64+1/64-1/128
=1/2+1/4+1/8+……+1/32+1/32-1/128
=1/2+1/4+1/8+1/16+1/16-1/128
=1/2+1/4+1/8+1/8-1/128
=1/2+1/4+1/4-1/128
=1/2+1/2-1/128
=1-1/128
=127/128
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