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(1+1/2)(1+1/2)^2(1+1/2)^4(1+1/2)^8(1+1/2)^16+(1/2)^31
具体过程····
人气:325 ℃ 时间:2019-12-01 13:50:11
解答
如果题目没有错误的话
(1+1/2)(1+1/2)^2(1+1/2)^4(1+1/2)^8(1+1/2)^16+(1/2)^31
=(1+1/2)^(1+2+4+8+16)+(1/2)^31
=(1+1/2)^31+(1/2)^31
=(3/2)^31+(1/2)^31
=(3×1/2)^31+(1/2)^31
=3^31×(1/2)^31+(1/2)^31
=(1/2)^31(3^31+1)
如果题目是这样的话
(1+1/2)(1+1/2)^2(1+1/2)^4(1+1/2)^8(1+1/2)^16(1/2)^31
=(1+1/2)^31(1/2)^31
=(3/2)^31(1/2)^31
=(3/4)^31
如果乘方在括号里面时,楼上的回答好像也有问题,下面好像应该为:
=(1-1/2^8)(1+1/2^8)(1+1/2^16)/(1-1/2) (看出来了吗,平方差公式一项项消掉了)
=(1-1/2^16)*(1+1/2^16)/(1-1/2)
=(1-1/2^32)/(1-1/2)
=(1-1/2^32)/(1/2)
=2(1-1/2^32)
=2-(1/2)^31
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