(1)
S(k²)=(Sk)²
k²a1+k²(k²-1)d/2=[ka1+k(k-1)d/2]²
a1=-4 d=2代入,整理,得
k²(k-3)=0
k>0,要等式成立,只有k=3
(2)
設(shè)Sn=an²+bn
a1=S1=a+b
S(k²)=(sk)²
a(k²)²+bk²=(ak²+bk)²
整理,得
(a²-a)k^4+2abk^3+(b²-b)k²=0
k>0,要對(duì)任意正整數(shù)k,等式恒成立,只有
a²-a=0
2ab=0
b²-b=0
解得a=0 b=0或a=0 b=1或a=1 b=0
a=0 b=0時(shí),Sn=0 數(shù)列是各項(xiàng)均為0的常數(shù)數(shù)列.
a=0 b=1時(shí),Sn=n,數(shù)列通項(xiàng)公式為an=1,數(shù)列是各項(xiàng)均為1的常數(shù)數(shù)列.
a=1 b=0時(shí),Sn=n²,an=2n-1 a1=2-1=1,數(shù)列是以1為首項(xiàng),2為公差的等差數(shù)列.