Let Dₖ=|1, 2 k, 2 k-1; n, n²+n+2, n²; n, n²+n, n²+n+2|. If Σk=1ⁿ Dₖ=96, then n is equal to _____________.

Solution

Maths sir · 4 steps

Only the first row changes with k, so add the determinants by adding that row alone.

1. Sum the first row. Rows 2 and 3 do not contain k. So Σ Dₖ is one determinant whose first row is the sum of the first rows. That row is Σ 1 = n, Σ 2k = n(n+1) and Σ (2k-1) = n².

   Σₖ₌₁ⁿ Dₖ = |n, n²+n, n²; n, n²+n+2, n²; n, n²+n, n²+n+2|

2. Subtract row 1. Do R₂ → R₂ - R₁ and R₃ → R₃ - R₁. Row 2 becomes (0, 2, 0) and row 3 becomes (0, 0, n+2), so the determinant is upper triangular.
3. Multiply the diagonal. The value is n × 2 × (n+2) = 2n(n+2). Setting 2n(n+2) = 96 gives n² + 2n - 48 = 0, that is (n+8)(n-6) = 0.
4. So. Since n counts terms, it must be positive. Reject n = -8, so n = 6.

The idea to keep. A sum of determinants that differ in only one row is a single determinant with that row summed. Then use row operations to create zeros before you expand.

Answer from the NTA final answer key. Solution by PrepWiser.