Question

在等差数列中,a2=4,a5=10求证(1)1/Sn=1/n-1/(n+1) (2)1／S1+1／S2+.+1／Sn

Answer 1

(1)a5-a2=3d=10-4,故

公差d=2,

a1=a2-d=4-2=2

1/sn=1/[n(a1+an)/2]=1/{n[2+2(n-1)+2]/2}=1/[(n+1)n]=1/n-1/(n+1)

(2)1／S1+1／S2+.+1／Sn=[1-1/(1+1)]+[1/2-1/(2+1)].+[1/n-1/(n+1)]

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

=1- 1/(n+1)