一道数学数列题，求详解已知数列｛
S6,S12-S6,S18-D12
S6=6a1+6(6-1)d/2=6a1+30d/2=6a1+15d
S12-S6=12a1+12*11d/2-6a1-30d/2=6a1+102d/2=6a1+51d
S18-S12=18a1+18*17d/2-12a1-12*11d/2=6a1+87d
S6+S18-S12=6a1+15d+6a1+87d=12a1+102d
2*s12=2*(6a1+51d)=12a1+102d=S6+S18-S12
S6,S12-S6,S18-S12成等差数列,
同理：
Sk,S2k-Sk,S3k-S2k
Sk=ka1+k(k-1)d/2
S2k-Sk=2ka1+...全部
S6,S12-S6,S18-D12
S6=6a1+6(6-1)d/2=6a1+30d/2=6a1+15d
S12-S6=12a1+12*11d/2-6a1-30d/2=6a1+102d/2=6a1+51d
S18-S12=18a1+18*17d/2-12a1-12*11d/2=6a1+87d
S6+S18-S12=6a1+15d+6a1+87d=12a1+102d
2*s12=2*(6a1+51d)=12a1+102d=S6+S18-S12
S6,S12-S6,S18-S12成等差数列,
同理：
Sk,S2k-Sk,S3k-S2k
Sk=ka1+k(k-1)d/2
S2k-Sk=2ka1+2k(2k-1)d/2-ka1-k(k-1)d/2=ka1+k(3k-1)d/2
S3k-S2k=3ka1+3k(3k-1)d/2-2ka1+2k(2k-1)d/2=ka1+k(5k-1)d/2
Sk+S3k-S2k=ka1+k(k-1)d/2+ka1+k(5k-1)d/2=2ka1+k(3k-1)d
2*(S2k-Sk)=2*(ka1+k(3k-1)d/2)=2ka1+k(3k-1)d
所以：
S6,S12-S6,S18-S12成等差数列
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