1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/(1+2+3+4.+2000)
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1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/(1+2+3+4.+2000)
![1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/(1+2+3+4.+2000)](/uploads/image/z/13506018-42-8.jpg?t=1%2F%EF%BC%881%2B2%EF%BC%89%2B1%2F%EF%BC%881%2B2%2B3%EF%BC%89%2B1%2F%EF%BC%881%2B2%2B3%2B4%EF%BC%89%2B.%2B1%2F%EF%BC%881%2B2%2B3%2B4.%2B2000%EF%BC%89)
1+2+...+n=n*(n+1)/2
1/(1+2+..+n)=2/[n*(n+1)]=2*[1/n-1/(n+1)]
原式
=2*(1/2-1/3)+2*(1/3-1/4)+...+2*(1/2000-1/2001)
=2*(1/2-1/3+1/3-1/4+...+1/2000-1/2001)
=2*(1/2-1/2001)
=1-2/2001
=1999/2001
1/(1+2+..+n)=2/[n*(n+1)]=2*[1/n-1/(n+1)]
原式
=2*(1/2-1/3)+2*(1/3-1/4)+...+2*(1/2000-1/2001)
=2*(1/2-1/3+1/3-1/4+...+1/2000-1/2001)
=2*(1/2-1/2001)
=1-2/2001
=1999/2001
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/(1+2+3+4.+2000)
如何计算:(1-1/2*2)(1-1/3*3)(1-1/4*4)(1-1/5*5)、、、(1-1/2000*2000)拜
1+2+3+4.+n(n+1)×1/2
200*(1-1/2)*(1-1/3)*(1-1/4)*.*(1-1/100)
(1) 1,1/3,1/5,1/7,1/9 (2)- 1/2*1,1/2*2,- 1/2*3,1/2*4,-1/2*5
一道奥数题,急救!(1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/
(1-2分之1)*(1-3分之1)*(1-4分之1)*...*(1-1999分之1)*(1-2000分之1) 等于
(1-2分之1)*(1-3分之1)*(1-4分之1)*...*(1-1999分之1)*(1-2000分之1) 等于 等于
(-1×1/2)+(-1/2×1/3)+(-1/3×1/4)+…+(-1/2008×1/2009)
怎么计算,(1-1/2)+(1/2-1/3)+(1/3-1/4)+.(1/1999-1/2000)
计算:[1+(-1/2)]+[1/2+(-1/3)]+[1/3+(-1/4)]+……+[1/1999+(-1/2000)
(-1/2)-(-1/3)-(+1/4)