sk=a1k+k(k-1)d/2=3k/2+k(k-1)/2=k(k+2)/2
sk^2=k^2(k^2+2)/2=(sk)^2=k^2(k+2)^2/4
2(k^2+2)=k^2+4k+4
k^2=4k
k=4（2）？sk=a1k+k(k-1)d/2=k[a1+d(k-1)/2]sk^2=k^2[a1+d(k^2-1)/2]=(sk)^2=k^2[a1+d(k-1)/2]^2a1+d(k^2-1)/2=a1^2+a1d(k-1)+d^2(k-1)^2/4k=1a1=a1^2, a1=0 or 1k=2, 3d/2=a1d+d^2/4, d=0 or 6-4a1a1=0, d=0 or 6a1=1, d=0 or 2d=0，為常數(shù)序列，a1=0,or 1 都滿足。d=2, a1=1, Sk=k^2, 也滿足d=6,a1=0, an=6(n-1), sn=3(n-1)n, sn^2=3n^2(n-1)^2, s(n^2)=3n^2(n^2-1),兩者不等。因此只有上面三種情況