已知数列an满足an=1/n（n+1）。求前n项和sn

更新时间：2018-08-11 18:15:06 125次访问

an=1/n(n+1)

=1/n - 1/(n+1)

故Sn=a1+a2+...+an

=(1-1/2)+(1/2-1/3)+...+[1/n- 1/(n+1)]

=1-1/(n+1)

=n/(n+1)

a(n)

=1/[n(n+1)]

=1/n-1/(n+1)

S(n)

=a(1)+a(2)+a(3)+......+a(n)

=(1-1/2)+(1/3-1/4)+(1/4-1/5)+......+[1/n-1/(n+1)]

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.....+1/n-1/(n+1)

=1-1/(n+1)

=(n+1)/(n+1)-1/(n+1)

=n/(n+1)