已知数列〔an〕是等差数列,sn是前n项之和,求证s6,s12_s6,s18-s12也成等差数列

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S6=(a1+a6)*6/2 =3(a1+a6)=6a1+15d
S12-S6=(a1+a12)*12/2-(a1+a6)*6/2
= 6(a1+a12)-3(a1+a6)
= 3a1+6a12-3a6
= 6a1+51d
同理
S18-S12=6a1+87d
2(S12-S6)=2*(6a1+51d )=12a1+102d
S18-S12+S6=6a1+87d+6a1+15d=12a1+102d
所以S6,S12-S6,S18-S12也成等差数列
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