(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16)+1/2^31
来源:百度知道 编辑:UC知道 时间:2024/07/06 20:54:59
简便算法!!!要有过程!!!
你又发了一遍么?
那我复制过来了
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16)这个式子左边乘以再除以(1-1/2)
则式子变为(1-1/2)(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16)/(1-1/2)
=(1-1/2^2)*(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16)/(1-1/2)
=(1-1/2^4)*(1+1/2^4)(1+1/2^8)(1+1/2^16)/(1-1/2)
=(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)*2+2^31
=2-1/2^16+2^31
我估计你最后一项是打错了
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
(1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100)
1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100)
1+1/1+2+1/1+2+3+...+1/1+2+3...+2000
1+1/1+2+1/1+2+3.........+1/1+2+3.....100
1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=?
1+1/2+1/3+.....+1/n
1+1/2+1/3+...+1/100
1-1/2+1/3-.....-1/10