①
后项比前项多8,公差为8:
S(N) = [9 + 9+8(N-1) ] *N / 2 = (8N+10)*N/2 = (4N+5)*N = 4N² + 5N
②
令 4N² + 5N = A²
(A + 2N)( A - 2N) = 5N
(A/N + 2)(A/N - 2) = 5
令A/N = T,
(T + 2)(T - 2) = 5,T² = 3,解得T = 3、-3（舍弃）
当T = 3时,A = 3N
4N² + 5N = 9N²
解得N = 1、0（舍弃）
综上,N = 1