(1+1/2)(1-1/2)(1+1/3)(1-1/3)........(1+1/99)(1-1/99)简便算法
来源:百度知道 编辑:UC知道 时间:2024/09/24 09:36:00
你设an=(1+1/n)(1-1/n)=(n-1)/n*(n+1)/n
求a2a3a4,,,,an(题目里n=99)
可以注意到:每一项和前面一项相乘都约去一项,剩下一下,
那么依次类推,就可以得到:
所有项都乘起来,就剩下第一项的1/2和最后一项(n+1)/n
那么
a2a3....an=(n+1)/2n
具体数字你再具体代入计算就好了,,这里n=99,那么
(1+1/2)(1-1/2)(1+1/3)(1-1/3)........(1+1/99)(1-1/99)
=(99+1)/99*2
=50/99
(3/2)*(1/2)*(4/3)*(2/3)......(100/99)*(98/99)
(3/2)*(4/3)*(5/4)...(99/98)*(100/99)*(1/2)*(2/3)*(3/4)...(97/98)*(98/99)
约分得(100/2)*(1/99)
=50/99
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
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)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
1+1/2+1+1/3+1+1/4+......+1/100=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)