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

an = a1+(n-1)d
Sn =(2a1+(n-1)d)n/2
S18-S12 = (2a1+17d)9 - (2a1+11d)6
= 6a1+87d
S12-S6 = (2a1+11d)6 - (2a1+5d)3
=6a1 +51d
S6 = (2a1+5d)3 = 6a1+15d

(S18-S12) = 6a1+87d -(6a1 +51d) =36d
S12-S6 =6a1 +51d - (6a1+15d) = 36d = (S18-S12)
=>S6,S12-S6,S18-S12成等差数列。

Answer 2

第一项为a,等差b
S6=6a+15b
S12=12a+66b
s18=18a+153b=3*(6a+51b)=3S12-3S6
2S12-2S6=S18-S12+S6
S6,S12-S6,S18-S12也成等差数列；

Answer 3

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.