Question

(1/1+2)+(1/1+2+3)+(1/1+2+3+4)+........+(1/1+2+3.....+20)怎么算？请大师写出步骤 ！我很着急，帮帮忙！

Answer 1

1+2+...+n=n(n+1)/2，所以1/(1+...+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
所以原式改写为2/(2*3)+2/(3*4)+...+2/(20/21)=2[(1/2-1/3)+(1/3-1/4)+...+(1/20-1/21)]
=2(1/2-1/21) =19/21

Answer 2

由Sn=n(a1+an)/2
得原式=2/2*3+2/3*4+2/4*5+2/5*6........2/20*21
=2[(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+1/5........+(1/20-1/21)]
=2(1/2-1/21)
=19/21