=（2/4+1/4）+（2/16+1/16）+（2/64+1/64）+1/128
=3/4+3/16+3/64+1/128
=3x(1/4+1/16+1/64)+1/128
=3x(16/64+4/64+1/64)+1/128
=3x(21/64)+1/128
=63/64+1/128
=126/128+1/128
=127/128
计算1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128怎么简便计算
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 =...
计算1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128怎么简便计算
法一：原式+1\/128=1 所以原式=127\/128 法二：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 ...
计算1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128怎么简便计算
原式=（1\/2+1\/4）+（1\/8+1\/16）+（1\/32+1\/64）+1\/128 =（2\/4+1\/4）+（2\/16+1\/16）+（2\/64+1\/64）+1\/128 =3\/4+3\/16+3\/64+1\/128 =3x(1\/4+1\/16+1\/64)+1\/128 =3x(16\/64+4\/64+1\/64)+1\/128 =3x(21\/64)+1\/128 =63\/64+1\/128 =126\/128+1\/128 =127...
1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128的巧算
=127\/128 1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128+1\/128-1\/128 因为每一项后面的分母都是前面的一半,所以你可以在原来的式子最后+1\/128,可以发现从后往前算,就是2个1\/128加起来变成1个1\/64,然后2个1\/64加起来变成1个1\/32,依次类推 ...
1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128[简便]?
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\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128等多少?简便计算
加第四项等于15\/16。第一项加第二项加第三项加第四项加第五项等于31\/32。从中可以发现他的答案就是分子比分母少一，∴然后加起来的答案就是127\/128。如果是高中的话，那么这道题就是一个等比数列。它的公比为1\/2。所以可根据公式1-q分之a1(1-q的n)∴最后的答案也是127\/128。
...1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128这三个简便运算
=15\/16 1\/2+1\/4+1\/8+1\/16+1\/32 =1\/2+(1\/2-1\/4)+(1\/4-1\/8)+(1\/8-1\/16)+(1\/16-1\/32)=1-1\/32 =31\/32 1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128 =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\/...
1\/2+1\/4+1\/8+1\/16+1\/32+1\/64简便计算的方法和思路?
简便计算1\/2+1\/4+1\/8+1\/16+1\/32+1\/64 解题思路：四则运算规则（按顺序计算、先算乘除后算加减，有括号先算括号，有乘方先算乘方）即脱式运算（递等式计算）需在该原则前提下进行 解题过程：1\/2+1\/4+1\/8+1\/16+1\/32+1\/64 =（4\/8+2\/8+1\/8）+（4\/64+2\/64+1\/64）=7\/8+7\/...
计算1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128怎么简便计算
=1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/128+1\/128-1\/128 =1\/2+1\/4+1\/8+1\/16+1\/32+1\/64+1\/64-1\/128 =1\/2+1\/4+1\/8+1\/16+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 =1-1\/128 =127\/128...
1\/2+1\/4+1\/8,加1\/16+1\/32+1\/64简便计算?
1\/64+1\/64)-1\/64 =1\/2+1\/4+1\/8+1\/16+(1\/32+1\/32)-1\/64 =1\/2+1\/4+1\/8+(1\/16+1\/16)-1\/64 =1\/2+1\/4+(1\/8+1\/8)-1\/64 =1\/2+(1\/4+1\/4)-1\/64 =1\/2+1\/2-1\/64 =1-1\/64 =63\/64 中间的过程可以略写的，只是为了让你方便理解，才一步步写出 ...
