Question

一道简便运算 数学 跪求呀 给分

1+1/(1+2)+1/(1+2+3）+......+1/(1+2+...+100) 来个过程呀

Answer 1

1/(1+2+...+n) = 2/(n)(n+1) = 2[1/n - 1/(n+1)]

1+1/3+1/6+1/10+1/15+。。。。。+1/1+2+3+。。。+n
=2 (1/1-1/2+1/2-1/3+1/3-1/4 + ...+1/n-1/n+1)
=2 [1-1/(n+1)]
=2n/(n+1)
参考答案：http://zhidao.baidu.com/question/86241458.html

Answer 2

1+1/(1+2)+1/(1+2+3）+......+1/(1+2+...+100)这是一个数列
an=1/[n(n+1)/2]=2/n(n+1)=2[1/n - 1/(n+1)]
1+1/(1+2)+1/(1+2+3）+......+1/(1+2+...+100)
=2[1/1-1/2+1/2-1/3+......+1/99 - 1/100+1/100-1/101]
=2[1-1/101]
=200/101

Answer 3

分母是等差数列，先用等差数列前n项和公式求出分母，然后用裂项相消法便可求解。具体的如下：
1+1/(1+2)+1/(1+2+3)+…+1/(1+2+…+100)
=2/2+2/2*(1+2)+2/3*(1+3)+…2/100*(1+100)
=2/1*2+2/2*3+2/3*4+…2/100*101
=2*(1/1*2+1/2*3+1/3*4+…1/100*101)
=2*(1-1/2+1/2-1/3+1/3-1/4+…+1/100-1/101)
=2*100/101
=200/101