Question

已知等差数列的前3项为a,4,3a,前n项和为Sn,Sk=2550

Answer 1

a+3a=2*4

4a=8

a=2

即a1=2

a2=4

a3=6

d=a2-a1=4-2=2

an=a1+(n-1)d

=2+2(n-1)

=2n

sn=(a1+an)*n/2

=(2+2n)*n/2

=n(n+1)

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

1/S1+1/S2+……1/Sn

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

=1-1/(n+1)

=n/(n+1)