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
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]^2
a1+d(k^2-1)/2=a1^2+a1d(k-1)+d^2(k-1)^2/4
k=1
a1=a1^2,a1=0 or 1
k=2,3d/2=a1d+d^2/4,d=0 or 6-4a1
a1=0,d=0 or 6
a1=1,d=0 or 2
d=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