Consider the sum (3)2 + (8)2+ (13)2+ ... + (15n - 2)2. m Use identities M k = NIH [m(m + 1)] and K =1 m [m(m + 1)(2m + 1)] to prove k=1 that (3)2+ (8)-+ (13)-+ ...+ (15n - 2)- = (n) (450n2 + 45n - 11)