Question

以知数列{An}是等差数列,{Sn}是其前n项的和,求证：S6,S12-S6,S18-S12 成等差数列.

Answer 1

S12-S6=a7+a8+a9+...+a12

(S12-S6)-s6=a7+a8+a9+...+a12-(a1+a2+...+a6)

=(a7-a1)+(a8-a2)+...+(a12-a6)

=6d+6d+...6d

=6d^2

S18-S12=a13+a14+...+a18

(S18-S12)-(S12-S6)=(a13-a7)+(a14-a8)+...(a18-a12)

=6d+6d+...+6d

=6d^2

S6,S12-S6,S18-S12 成等差数列,且公差为6d^2.

推倒：Sn,S2n-Sn,S3n-S2n成等差数列,且公差为nd^2.