(1+1/2+1/3+1/4)*(1/2+111/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)=简便方
来源:百度知道 编辑:UC知道 时间:2024/05/26 10:52:47
用简便方法,并有过程,谢了!!!!
1/2+1/3+1/4=t
原式=(1+t)(1/5+t)-(1+1/5+t)t
=1/5+6t/5+t^2-6t/5-t^2
=1/5
设:a=(1+1/2+1/3+1/4)
原式
=a(a-1+1/5)-(a+1/5)(a-1)
=a^2-a+a/5-a^2-a/5+a+1/5
=1/5
设1+1/2+1/3+1/4+1/5=x,
原式=(x-1/5)(x-1)-x(x-1-1/5)
=x^2-6x/5+1/5-x^2+6/5x
=1/5
(1+1/2+1/3+1/4)*(1+1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)
=(1+1/2+1/3+1/4+1/5)*(1+1/2+1/3+1/4-(1/2+1/3+1/4)
=1+1/2+1/3+1/4+1/5=137/60
设1+1/2+1/3+1/4+1/5=x,
原式=(x-1/5)(x-1)-x(x-1-1/5)
=x^2-6x/5+1/5-x^2+6/5x
=1/5
111/3?
(1+1/2+1/3+1/4)×
(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)
1/1+1/2+1/3+1/4+。。。。+1/N 是多少
1/1+1/2+1/3+1/4+......1/2002=?
1-1/2+1/3-1/4+........1/99-1/100
求和Sn=1+(1+1/2)+(1+1/2+1/4)+....+[1+1/2+1/4.....+1/2^(n-1)]
数列 1+(1+1/2)+(1+1/2+1/4)+..............=?