UPSC NDA II 2024 Mathematics Paper

Concept:The determinant of a matrix can be simplified by factoring out common column factors, and summation over determinants is evaluated using standard sum formulas.Explanation:Given $D_n = \begin{vmatrix} n & 20 & 30 \\ n^2 & 40 & 50 \\ n^3 & 60 & 70 \end{vmatrix}$.Factor out common terms: $20$ from the second column and $10$ from the third column.This gives $D_n = 200 \begin{vmatrix} n & 1 & 3 \\ n^2 & 2 & 5 \\ n^3 & 3 & 7 \end{vmatrix}$.Expand the determinant along the first column:$\begin{vmatrix} n & 1 & 3 \\ n^2 & 2 & 5 \\ n^3 & 3 & 7 \end{vmatrix} = n(2\cdot7 - 5\cdot3) - n^2(1\cdot7 - 3\cdot3) + n^3(1\cdot5 - 3\cdot2)$.Calculate each term: $n(14-15) = -n$, $-n^2(7-9) = 2n^2$, $n^3(5-6) = -n^3$.So the determinant simplifies to $-n + 2n^2 - n^3$.Thus $D_n = 200(-n + 2n^2 - n^3)$.Now compute $\sum_{n=1}^{4} D_n = 200 \sum_{n=1}^{4} (-n + 2n^2 - n^3)$.Use summation formulas: $\sum_{n=1}^{4} n = \frac{4\cdot5}{2} = 10$, $\sum_{n=1}^{4} n^2 = \frac{4\cdot5\cdot9}{6} = 30$, $\sum_{n=1}^{4} n^3 = \left(\frac{4\cdot5}{2}\right)^2 = 100$.Substitute: $\sum_{n=1}^{4} (-n + 2n^2 - n^3) = -10 + 2\cdot30 - 100 = -10 + 60 - 100 = -50$.Therefore, $\sum_{n=1}^{4} D_n = 200 \times (-50) = -10000$.Answer:$-10000$ (Option A).