Question

设无穷等差数列{An}的前n项和为Sn,求所有的无穷等差数列{An},使得对于一切正整数k都有S(k^2)=(Sk)^2

Answer 1

由题设 令k =1 有S1 = (S1)^2 即a1 = a1^2

得a1 =0 或者a1 = 1

当a1=0时,Sn = n(n-1)d/2 由题设有

k^2(k^2-1)d/2 = (k(k-1)d/2)^2 = k^2(k-1)^2d^2/4

(k^2-1)d = (k-1)^2d^2/2

(k+1)d=(k-1)d^2/2

如果d=0则等式成立,否则d=2(k+1)/(k-1)与k相关,不成立

所以当a1 =0 时,d = 0满足条件

当k = 1时,Sn = n + n(n-1)d/2 = n(1+(n-1)d/2)

k^2(1+(k^2-1)d/2) = k^2(1+(k-1)d/2)^2

1+(k^2-1)d/2 = (1+(k-1)d/2)^2

1+(k^2-1)d/2 = 1 + (k-1)d + (k-1)^2d^2/4

(k^2-1)d/2 = (k-1)d + (k-1)^2d^2/4

(k+1)(k-1)d/2 = (k-1)d + (k-1)^2d^2/4

同除以k-1得

(k+1)d/2 = d + (k-1)d^2/4

(k+1)d/2 -d = (k-1)d^2/4

(k-1)d/2 = (k-1)d^2/4

同除以k-1 得

d/2 = d^2/4

得2d = d^2

d = 0 或者2

所以当a1 = 1 时,d =0 或者2 满足条件

综上,符合条件得数列通项为：

an = 0

或者an = 1

或者an = 1 + 2(n-1)