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 _________ . [2023]

Question

$Let D_{k} =$ $| \begin{matrix} 1 & 2 k & 2 k - 1 \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$ . $If \sum_{k = 1}^{n} D_{k} = 96, then n is equal to$ _________ . [2023]

Smart Achievers · Accepted Answer

(6)

$Given, D_{k} =$ $| \begin{matrix} 1 & 2 k & 2 k - 1 \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$

$∴$ $\sum_{k = 1}^{n} D_{k} =$ $| \begin{matrix} \sum_{k = 1}^{n} 1 & \sum_{k = 1}^{n} 2 k & \sum_{k = 1}^{n} (2 k - 1) \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$

$Apply \sum_{k = 1}^{n} 2 k = \frac{2 n (n + 1)}{2} = n (n + 1)$

$\sum_{k = 1}^{n} (2 k - 1) = \sum_{k = 1}^{n} 2 k - \sum_{k = 1}^{n} 1 = \frac{2 n (n + 1)}{2} - n = n^{2} + n - n = n^{2}$

$∴$ $\sum_{k = 1}^{n} D_{k} =$ $| \begin{matrix} n & n^{2} + n & n^{2} \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$

$Apply R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}$

$\Rightarrow$ $| \begin{matrix} n & n^{2} + n & n^{2} \\ 0 & 2 & 0 \\ 0 & 0 & n + 2 \end{matrix} |$ $= 96$

$\Rightarrow n (2 (n + 2) - 0) = 96 \Rightarrow n (n + 2) = 48$

$\Rightarrow n^{2} + 2 n - 48 = 0 \Rightarrow n^{2} + 8 n - 6 n - 48 = 0 \Rightarrow n (n + 8) - 6 (n + 8) = 0 \Rightarrow (n - 6) (n + 8) = 0$

$\Rightarrow n = 6, - 8$

$Since n cannot be negative, ∴$ $n = 6$

Solution

Q.
$Let D_{k} =$ $| \begin{matrix} 1 & 2 k & 2 k - 1 \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$ . $If \sum_{k = 1}^{n} D_{k} = 96, then n is equal to$ _________ . [2023]

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Solution

Q. Let Dk=|12k2k-1nn2+n+2n2nn2+nn2+n+2|. If ∑k=1nDk=96, then n is equal to _________ . [2023]

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Q.
$Let D_{k} =$ $| \begin{matrix} 1 & 2 k & 2 k - 1 \\ n & n^{2} + n + 2 & n^{2} \\ n & n^{2} + n & n^{2} + n + 2 \end{matrix} |$ . $If \sum_{k = 1}^{n} D_{k} = 96, then n is equal to$ _________ . [2023]