Question

已知数列{an}的通项公式为an=1/1+2+3+...+n求这个数列的前n项和sn

Answer 1

an=1/(1+2+3+...+n)
=1/[n(n+1)]/2
=2/n(n+1)
Sn=2{1/1*2+1/2*3+1/3*4+...+1/n(n+1)}
=2*{1-1/2+1/2-1/3+1/3-1/4...+1/(n-1)+1/n+1/n-1/(n+1)}
=2*{1-1/(n+1)}
=2n/(n+1)

Answer 2

an=1/(n+1)*n/2=2/(n+1)*n=2(1/n-1/n+1)
Sn=2(1-1/2+1/2-1/3+1/3-1/4……-1/n+1)
=2(1-1/n+1)
=2n/n+1

Answer 3

解：
an=1/(1+2+3+...+n)=1/[n(n+1)/2]=2/[n(n+1)]=2[1/n-1/(n+1)]
Sn=a1+a2+a3+...+an
=2[1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)