Let S k = 1 + 2 + 3 + … + k k . If S 1 2 + S 2 2 + … + S 10 2 = 5 12 A , then A is equal to :

Step-by-step solution

S k = 1 + 2 + 3 + … + k k = k k + 1 2 k = k + 1 2   . . . . . . . . . i ∴   S 1 2 + S 2 2 + . . . . + S 10 2 = 1 + 1 2 2 + 2 + 1 2 2 + . . . + 10 + 1 2 2 (using equation i ) = 2 2 2 + 3 2 2 + . . . + 11 2 2 = 1 2 2 2 2 + 3 2 + . . . + 11 2 = 1 4 1 2 + 2 2 + 3 2 + . . . + 11 2 - 1 2 = 1 4 11 × 12 × 23 6 - 1 (using the identity ∑ n 2 n = 1 n = n = n n + 1 2 n + 1 6 ) = 1 4 × 505 Comparing with the given condition in question, we get 5 12 A = 505 4 ⇒ A = 505 × 3 5