Question

已知等差数列前三项为a,4,3a,前n项和为Sn,Sk=2550.

Answer 1

a+3a=4*2

a=2

d=4-2=2

an=2+2(n-1)

Sn=[2+2+2(n-1)]*n/2=n(n+1)

1/S +1/S2 +.+1/Sn

=1/1*2+1/2*3+.+1/n(n+1)

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

=1-1/(n+1)

=n/(n+1)