Q18 / 120 · NDA 2024 (Phase 2) · Mathematics
If $\mathrm{D}_n=\left|\begin{array}{lll}n & 20 & 30 \\ n^2 & 40 & 50 \\ n^3 & 60 & 70\end{array}\right|$ then what is the value of $\sum_{n=1}^4 \mathrm{D}_n$ ?
- Correct answerA$-10000$
- B$-10$
- C$10$
- D$10000$
Solution
Given,
$\begin{aligned} & D_n=\left|\begin{array}{lll}n & 20 & 30 \\ n^2 & 40 & 50 \\ n^3 & 60 & 70\end{array}\right|=(20)(10)\left|\begin{array}{lll}n & 1 & 3 \\ n^2 & 2 & 5 \\ n^3 & 3 & 7\end{array}\right| \\ & \Rightarrow D_n=200\left[n(14-15)-1\left(7 n^2-5 n^3\right)+3\left(3 n^2-2 n^3\right)\right] \\ & \Rightarrow D_n=200\left(-n+2 n^2-n^3\right)\end{aligned}$
$\begin{aligned}
& \text { Now, } \sum_{n=1}^4 D_n=200 \sum_{n=1}^4\left(-n+2 n^2-n^3\right) \\
& \because \sum n=\frac{n(n+1)}{2}, \sum n^2=\frac{n(n+1)(2 n+1)}{6} \\
& \sum n^3=\left[\frac{n(n+1)}{2}\right]^2
\end{aligned}$
Hence,
$\begin{aligned}
& \sum_{n=1}^4 D_n=200\left[-\frac{(4)(5)}{2}+2 \cdot \frac{(4)(5)(9)}{6}-\left(\frac{(4)(5)}{2}\right)^2\right] \\
& =200(-50)=-10000
\end{aligned}$
Hence, the correct answer is option: 'A'.