Question

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

Answer 1

因为

S(6)=6a(1)+6×5d/2=6a(1)+15d

S(12)=12a(1)+66d

S(18)=18a(1)+153d

而

S(12)-S(6)=6a(1)+51d

S(18)-S(12)=6a(1)+87d

可见,S(6)、S(12)-S(6)、S(18)-S(12)之间满足

2[S(12)-S(6)]=S(6)+[S(18)-S(12)]

即三者成等差数列.

实际上,上面的结论可推广到S(n)、S(2n)-S(n)、S(3n)-S(2n)的情形.