If Dk=∣∣ ∣ ∣∣1nn2kn2+n+1n2+n2k−1n2n2+n

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If

D_{k} = ∣ ∣∣∣ ∣ \begin{matrix} 1 & n & n 2 k & n^{2} + n + 1 & n^{2} + n 2 k - 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣∣∣ ∣

and

\sum_{k = 1}^{n} D_{k} = 56

, then n equals

Options:

A . 4

B . 6

C . 8

D . None of these

Answer: Option D
:
D

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

= ∣ ∣∣∣ ∣ \begin{matrix} n & n & n n^{2} + n & n^{2} + n + 1 & n^{2} + n n^{2} & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣∣∣ ∣ = 56

=Applying

C_{2} \to C_{2} \to C_{1} \to C_{3} \to C_{3} - C_{1}

, then

∣ ∣∣ ∣ \begin{matrix} n & 0 & 0 n^{2} + n & 1 & 0 n & 0 & n + 1 \end{matrix} ∣ ∣∣ ∣ = 56 \Rightarrow n (n + 1) = 56 = 7 \times 8 \Rightarrow n = 7

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