Question:
Let \(\mathrm{A}_{1}=[1]\), \(\mathrm{A}_{2}=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]\), \(\mathrm{A}_{3}=\left[\begin{array}{ccc}6 & 7 & 8 \\ 9 & 10 & 11 \\ 12 & 13 & 14\end{array}\right], \ldots\) Then the sum of the diagonal elements of \(\mathrm{A}_{20}\) is ________.
Let \(\mathrm{A}_{1}=[1]\), \(\mathrm{A}_{2}=\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]\), \(\mathrm{A}_{3}=\left[\begin{array}{ccc}6 & 7 & 8 \\ 9 & 10 & 11 \\ 12 & 13 & 14\end{array}\right], \ldots\) Then the sum of the diagonal elements of \(\mathrm{A}_{20}\) is ________.
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Find the first entry of \(A_{20}\) using the sum of squares of orders of earlier matrices; the diagonal entries then form an AP with common difference 21.
Updated On: Nov 12, 2026
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Correct Answer: 53410
Solution and Explanation
\(A_n\) is an \(n\times n\) matrix filled with consecutive integers. Before \(A_{20}\) the matrices \(A_1,\dots,A_{19}\) use \(1^2+2^2+\dots+19^2=\frac{19\cdot 20\cdot 39}{6}=2470\) numbers (1 to 2470).
So \(A_{20}\) starts at 2471 and its entries run 2471, 2472, ..., 2870 (400 numbers) row by row.
The diagonal entry in row i is at position \((i-1)\cdot 20+(i-1)=21(i-1)\) from the start, i.e. \(2471+21(i-1)\), for \(i=1,\dots,20\).
Sum \(=20\cdot 2471+21(0+1+\dots+19)=49420+21\cdot 190=49420+3990=53410\).
Answer: 53410
So \(A_{20}\) starts at 2471 and its entries run 2471, 2472, ..., 2870 (400 numbers) row by row.
The diagonal entry in row i is at position \((i-1)\cdot 20+(i-1)=21(i-1)\) from the start, i.e. \(2471+21(i-1)\), for \(i=1,\dots,20\).
Sum \(=20\cdot 2471+21(0+1+\dots+19)=49420+21\cdot 190=49420+3990=53410\).
Answer: 53410
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