If D k - 1 n n- 2 k n 2- n -1 n 2- n- 2 k -1 n 2 n 2- n -1 - and - k -1n D k -56- then n equalsA- 4B- 6C-D- None of the above

The correct option is D None of the above nbsp;∑^n_k=1 D_k = 56 ⇒∑^n_k=1 amp; n amp; n nbsp;∑^n_k=1 2k amp; n^2 + n + 1 amp; n^2 + n nbsp; ...

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Evaluation of a Determinant

If D k =|[ ...

Question

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 $n \sum k = 1 D_{k} = 56,$ then $n$ equals

4

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6

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8

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None of the above

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Solution

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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} ∣ ∣ ∣ ∣ ∣

n \sum k = 1 D_{k} = 56

D_{k} = |\begin{matrix} 1 & n & n \\ 2 k & n^{2} + n + 2 & n^{2} + n \\ 2 k - 1 & n^{2} & n^{2} + n + 2 \end{matrix}| and \sum_{k = 1}^{n} D_{k} = 48

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} ∣ ∣ ∣ ∣ ∣

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

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

n \sum k = 1 D_{k} = 48

△_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & n & n 2 r & n^{2} + n + 1 & n^{2} + n 2 r - 1 & n^{2} & n^{2} + n + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

n \sum r = 1 △_{r} = 56

n

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