If Delta_k=left|begin{array}{ccc}1 & 0 & 0 0 | vedclass

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(D) Given $\Delta_k = \left|\begin{array}{ccc} 1 & 0 & 0 \ 0 & k & k-1 \ 0 & k-1 & k \end{array}ight|$.Expanding along the first row,we get:$\Delt

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$400$

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(D) Given $\Delta_k = \left|\begin{array}{ccc} 1 & 0 & 0 \ 0 & k & k-1 \ 0 & k-1 & k \end{array}ight|$.Expanding along the first row,we get:$\Delta_k = 1 \cdot (k \cdot k - (k-1) \cdot (k-1)) - 0 + 0$$\Delta_k = k^2 - (k-1)^2$Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have $\Delta_k = (k - (k-1))(k + (k-1)) = 1 \cdot (2k-1) = 2k-1$.We need to find the sum $S = \sum_{k=1}^{20} \Delta_k = \sum_{k=1}^{20} (2k-1)$.This is the sum of the first $20$ odd numbers,which is given by $n^2$ where $n=20$.$S = 20^2 = 400$.Alternatively,using the telescoping sum property:$\Delta_k = k^2 - (k-1)^2$$\sum_{k=1}^{20} \Delta_k = (1^2 - 0^2) + (2^2 - 1^2) + (3^2 - 2^2) + \ldots + (20^2 - 19^2) = 20^2 - 0^2 = 400$.

If $\Delta_k=\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & k & k-1 \\ 0 & k-1 & k\end{array}\right|$,then $\Delta_1+\Delta_2+\ldots+\Delta_{20}$ is equal to

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