If D_n=| array lll n & 20 & 30 n^2 & 40 & 50 n^3 & 60 & 70 array | then what is the value of ∑_ n=1 ^4 D_n ?

Correct answer: A. -10000 Given, aligned & D_n=| array lll n & 20 & 30 n^2 & 40 & 50 n^3 & 60 & 70 array |=(20)(10)| array lll n & 1 & 3 n^2 & 2 & 5 n^3 & 3 & 7 array | & D_n=200[n(14-15)-1(7 n^2-5 n^3)+3(3 n^2-2 n^3)] & D_n=200(-n+2 n^2-n^3) aligned aligned & Now, ∑_ n=1 ^4 D_n=200 ∑_ n=1 ^4(-n+2 n^2-n^3) & ∑ n=(n(n+1))/(2), ∑ n^2=(n(n+1)(2 n+1))/(6) & ∑ n^3=[(n(n+1))/(2)]^2 aligned Hence, aligned & ∑_ n=1 ^4 D_n=200[-((4)(5))/(2)+2 · ((4)(5)(9))/(6)-(((4)(5))/(2))^2] & =200(-50)=-10000 aligned

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Determinants

NDA 2024 · Mathematics · Determinants

If D_n=| array lll n & 20 & 30 n^2 & 40 & 50 n^3 & 60 & 70 array |...

\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

Options

A $-10000$
B $-10$
C $10$
D $10000$

Correct answer

A. $-10000$

Step-by-step 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}

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