a(n) = 3/2 + (n-1) = n + 1/2 = (2n+1)/2.
s(n) = n(n+1)/2 + n/2 = n(n+2)/2,
s[k^2] = k^2(k^2 + 2)/2,
[s(k)]^2 = [k(k+2)/2]^2 = k^2(k+2)^2/4,
s[k^2] = k^2(k^2 + 2)/2 = [s(k)]^2 = k^2(k+2)^2/4,
2(k^2 + 2) = (k+2)^2 = k^2 + 4k + 4,
0 = k^2 - 4k = k(k-4),
k=4.
----------------
a(n) = a + (n-1)d.
s(n) = na + n(n-1)d/2.
s[k^2] = ak^2 + k^2(k^2-1)d/2.
[s(k)]^2 = [ak + k(k-1)d/2]^2.
s[k^2] = ak^2 + k^2(k^2 - 1)d/2 = [s(k)]^2 = [ak + k(k-1)d/2]^2 = k^2[a + (k-1)d/2]^2,
a + (k^2 - 1)d/2 = [a + (k-1)d/2]^2 = [a-d/2 + kd/2]^2 = (a-d/2)^2 + (a-d/2)dk + (d/2)^2k^2,
0 = k^2(d/2 - d^2/4) - k(a-d/2)d + (a-d/2) - (a-d/2)^2
要使得上面等式恒成立,
則,
0 = d^2/4 - d/2 = (d/4)(d-2),d=2.
0 = (a-d/2)d,a = d/2 = 1.
0 = (a-d/2)-(a-d/2)^2 成立.
因此,a(n) = 1 + 2(n-1) = 2n-1.
只有無窮等差數(shù)列{a(n) = 2n-1}使得對于一切正整數(shù)k,都有s(k^2) = [s(k)]2成立.