Let S K = 1 + 2 + . . . + K K and ∑ j = 1 n S 2 j = n A B n 2 + C n + D where A , B , C , D ∈ N and A Has least value then

Step-by-step solution

Given that S K = 1 + 2 + . . . + K K We know that 1 + 2 + 3 . . . . + n = n n + 1 2 ⇒ S K = K K + 1 2 K = K + 1 2 Now, ∑ j = 1 n S j 2 = ∑ j = 1 n 1 4 2 2 + 3 2 + 4 2 . . . n + 1 2 = ∑ j = 1 n 1 4 1 2 + 2 2 + 3 2 + 4 2 . . . n + 1 2 - 1 2 ⇒ 1 4 n + 1 n + 2 2 n + 3 6 - 1 = 1 4 n 2 + 3 n + 2 2 n + 3 - 6 6 = 1 4 2 n 3 + 6 n 2 + 4 n + 3 n 2 + 9 n + 6 - 6 6 = 1 4 2 n 3 + 9 n 2 + 13 n 6 = n 24 2 n 2 + 9 n + 13 On comparing this with ∑ j = 1 n S 2 j = n A B n 2 + C n + D we get, A = 24