<p>The sum of number of elements of \(m_{n-1}\) matrices gives the last element of \(m_{n-1}\) matrix. The last element of \(m_{n-1}\) matrix is \(1^2+2^2+3^2+\cdots+(n-1)^2 = \dfrac{n(n-1)(2n-1)}{6}\). The first element of \(m_{10}\) matrix is \(\dfrac{10 \times 9 \times 19}{6}+1 = 286\). The common difference in diagonal elements of \(m_n\) matrix is \(n+1\); in \(m_{10}\), the common difference is 11. Find the sum of diagonal elements of \(m_{10}\).</p>
Step-by-Step Solution
Key Concept: Recognize that the diagonal elements of m_n form an arithmetic progression with first term = (n(n-1)(2n-1))/6 + 1 and common difference = n+1. Use the AP sum formula S = (number of terms/2)(2a + (k-1)d) where k is the number of diagonal elements.
<p><strong>Step 1:</strong> For m_10 matrix, identify the parameters of the arithmetic progression formed by diagonal elements.</p><p>• First diagonal element (a) = (10 × 9 × 19)/6 + 1 = 1710/6 + 1 = 285 + 1 = <strong>286</strong></p><p>• Common difference (d) = 10 + 1 = <strong>11</strong></p><p>• Number of diagonal elements (k) = <strong>10</strong></p><p><strong>Step 2:</strong> The diagonal elements form an AP: 286, 297, 308, ..., (10th term)</p><p><strong>Step 3:</strong> Find the 10th diagonal element using a_k = a + (k-1)d</p><p>a_10 = 286 + (10-1)(11) = 286 + 99 = <strong>385</strong></p><p><strong>Step 4:</strong> Apply the sum formula for AP: S = (k/2)(first term + last term)</p><p>S = (10/2)(286 + 385) = 5 × 671 = <strong>3355</strong></p><p>∴ Answer: <strong>3355</strong></p>
Correct Answer: 3355