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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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Verified Answer
The correct answer is:
400
We have,
$$
\begin{aligned}
\Delta_K & =\left|\begin{array}{ccc}
1 & 0 & 0 \\
0 & k & k-1 \\
0 & k-1 & k
\end{array}\right| \\
\Delta_K & =k^2-(k-1)^2 \\
\Delta_1+\Delta_2+\Delta_3 & +\ldots+\Delta_{20} \\
\therefore \quad & =\left(1^2-0^2\right)+\left(2^2-1^2\right)+\ldots\left(20^2-19^2\right) \\
& =20^2=400
\end{aligned}
$$
$$
\begin{aligned}
\Delta_K & =\left|\begin{array}{ccc}
1 & 0 & 0 \\
0 & k & k-1 \\
0 & k-1 & k
\end{array}\right| \\
\Delta_K & =k^2-(k-1)^2 \\
\Delta_1+\Delta_2+\Delta_3 & +\ldots+\Delta_{20} \\
\therefore \quad & =\left(1^2-0^2\right)+\left(2^2-1^2\right)+\ldots\left(20^2-19^2\right) \\
& =20^2=400
\end{aligned}
$$
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