1/(2^2-1)+1/(4^2-1)+1/(6^2-1)+...+1/9100^2-1)=?
来源:百度知道 编辑:UC知道 时间:2024/06/10 23:10:31
例题:求和:1/(2^2-1)+1/(4^2-1)+1/(6^2-1)+...+1/[(2n)^2-1]
解答:
每项都可以分解的
1/(2^2-1)=1/2(1-1/2)
1/(4^2-1)=1/2(1/3-1/5)
总的就=1/2(1-1/2+1/3-1/5+1/5-1/7+1/7-1/9+...-1/(2n+1))
整理下=5/12+1/(4n+2)
原式=1/2*(1-1/3+1/3-1/5+1/5-1/7+……+1/9099-1/9101)
=1/2*(1-1/9101)
=4550/9101
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+1/2^2)(1+1/2^4)(1+1/2^8)
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)
1/2-1/2=?
3/2=2+1/1*2=1/1+1/2
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)