<p><strong>15.</strong> Consider \(a_{33}\) it is the biggest number in the \(3 \times 3\) sub matrix formed by top left corner so minimum value of \(a_{33}\) is 9. Now again consider \(a_{33}\) it is the smallest number in \((n-3)(n-3)\) sub matrix formed by bottom left corner of the main matrix. If \(b_j\) denotes the number of elements in the set \(\{a_{jj} : j^2 \leq a_{jj} \leq n^2 - (n-j+1)^2 + 1\}\), find \(\sum_{j=1}^{n} b_j\).</p>
Step-by-Step Solution
Key Concept: For each diagonal element a_{jj}, we need to count how many integers satisfy j² ≤ a_{jj} ≤ n² - (n-j+1)² + 1, then sum these counts across all j from 1 to n. This requires simplifying the upper bound expression and carefully determining the valid range for each j.
<p><strong>Step 1: Simplify the upper bound constraint</strong></p><p>The upper bound is n² - (n-j+1)² + 1.</p><p>Expanding: (n-j+1)² = n² - 2n(j-1) + (j-1)²</p><p>Therefore: n² - (n-j+1)² + 1 = n² - [n² - 2n(j-1) + (j-1)²] + 1</p><p>= 2n(j-1) - (j-1)² + 1</p><p>= 2nj - 2n - j² + 2j - 1 + 1</p><p>= 2nj - 2n - j² + 2j</p></p><p><strong>Step 2: Count valid integers for each j</strong></p><p>For diagonal element a_{jj}, we need: j² ≤ a_{jj} ≤ 2nj - 2n - j² + 2j</p><p>The number of integers b_j in this range is:</p><p>b_j = (2nj - 2n - j² + 2j) - j² + 1</p><p>= 2nj - 2n - 2j² + 2j + 1</p><p>= 2nj + 2j - 2j² - 2n + 1</p></p><p><strong>Step 3: Compute the sum ∑_{j=1}^{n} b_j</strong></p><p>∑_{j=1}^{n} b_j = ∑_{j=1}^{n} (2nj + 2j - 2j² - 2n + 1)</p><p>= 2n∑_{j=1}^{n} j + 2∑_{j=1}^{n} j - 2∑_{j=1}^{n} j² - 2n² + n</p></p><p><strong>Step 4: Apply summation formulas</strong></p><p>Using: ∑_{j=1}^{n} j = n(n+1)/2 and ∑_{j=1}^{n} j² = n(n+1)(2n+1)/6</p><p>= 2n · n(n+1)/2 + 2 · n(n+1)/2 - 2 · n(n+1)(2n+1)/6 - 2n² + n</p><p>= n²(n+1) + n(n+1) - n(n+1)(2n+1)/3 - 2n² + n</p><p>= n(n+1)(n+1) - n(n+1)(2n+1)/3 - 2n² + n</p><p>= n(n² + 2n + 1) - n(n+1)(2n+1)/3 - 2n² + n</p><p>= n³ + 2n² + n - n(2n² + 3n + 1)/3 - 2n² + n</p><p>= n³ + 2n - n(2n² + 3n + 1)/3</p><p>= [3n³ + 6n - 2n³ - 3n² - n]/3</p><p>= [n³ - 3n² + 5n]/3</p><p>= n(n² - 3n + 5)/3</p><p><strong>∴ Answer:</strong> n(n²-3n+5)/3</p>
Correct Answer: n(n²-3n+5)/3