<p>The value of \(\displaystyle\sum_{r=16}^{30}(r+2)(r-3)\) is equal to:</p>
Step-by-Step Solution
Key Concept: Expand (r+2)(r-3) = r² - r - 6, then use summation formulas for ∑r², ∑r, and ∑1 over the given range. Alternatively, use telescoping or shift the index to simplify calculations.
<p><strong>Step 1:</strong> Expand the expression:</p><p>(r+2)(r-3) = r² - 3r + 2r - 6 = r² - r - 6</p><p><strong>Step 2:</strong> Split the summation:</p><p>∑(r=16 to 30) (r² - r - 6) = ∑(r=16 to 30) r² - ∑(r=16 to 30) r - 6∑(r=16 to 30) 1</p><p><strong>Step 3:</strong> Use the technique of subtracting partial sums:</p><p>∑(r=16 to 30) r² = ∑(r=1 to 30) r² - ∑(r=1 to 15) r² = [30(31)(61)/6] - [15(16)(31)/6] = 9455 - 1240 = 8215</p><p>∑(r=16 to 30) r = ∑(r=1 to 30) r - ∑(r=1 to 15) r = [30(31)/2] - [15(16)/2] = 465 - 120 = 345</p><p>∑(r=16 to 30) 1 = 30 - 16 + 1 = 15</p><p><strong>Step 4:</strong> Combine results:</p><p>∑(r=16 to 30) (r² - r - 6) = 8215 - 345 - 6(15) = 8215 - 345 - 90 = 7780</p><p>∴ Answer: A</p>
Correct Answer: A