Question

数学问题：1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+…+100)=？

1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+…+1/(1+2+3+…+100)=？
谢谢大家，动动脑筋吧！

Answer 1

=2*(1/2*3+1/3*4+……+1/(n+1)*(n+2))
=2(1/2-1/3+1/3-1/4+1/4……+1/n+1-1/n+2)
=2(1/2-1/n+2)
n=99
原=1-2/101=99/101

Answer 2

利用等差数列求和公式表示分母，然后用裂项法解决

Answer 3

=2/[（1+2）*2]+2/[(1+3)*3]+……+2/[(1+100)*100]
=2*[1/2-1/3+1/3-1/4+……+1/100-1/101]
=2*(1/2-1/101)
=99/101

Answer 4

你好2~~~！！