Let D k = 1 2 k 2 k - 1 n n 2 + n + 2 n 2 n n 2 + n n 2 + n + 2 . If ∑ k = 1 n D k = 96 , then n is equal to _________.

Step-by-step solution

Given that D k = 1 2 k 2 k - 1 n n 2 + n + 2 n 2 n n 2 + n n 2 + n + 2 ⇒ ∑ k = 1 n D k = ∑ 1 2 ∑ k 2 ∑ k - ∑ 1 n n 2 + n + 2 n 2 n n 2 + n n 2 + n + 2 ⇒ ∑ k = 1 n D k = n 2 n n + 1 2 2 n n + 1 2 - n n n 2 + n + 2 n 2 n n 2 + n n 2 + n + 2 ⇒ ∑ k = 1 n D k = n n n + 1 n 2 n n 2 + n + 2 n 2 n n 2 + n n 2 + n + 2 R 1 → R 1 - R 2 and R 2 → R 2 - R 3 ⇒ ∑ k = 1 n D k = 0 - 2 0 0 2 - n - 2 n n 2 + n n 2 + n + 2 Expand the determinant along