Sequences & Series
Sum of Series
Grade 11
Question:
<p>Let \(a, b, c \in \mathbb{R}\). If \(f(x) = ax^2 + bx + c\) is such that \(a + b + c = 3\) and \(f(x + y) = f(x) + f(y) + xy\), \(\forall\, x, y \in \mathbb{R}\), then \(\displaystyle\sum_{n=1}^{10} f(n)\) is equal to</p>
<p>255</p>
<p>330</p>
<p>165</p>
<p>190</p>
Step-by-Step Solution
Key Concept: Use the functional equation f(x+y) = f(x) + f(y) + xy to determine coefficients by substituting specific values (y=1, then x=y=1), then combine with the constraint a+b+c=3 to find f(x) explicitly.
<p><strong>Step 1:</strong> Expand f(x+y) using f(x) = ax² + bx + c:</p><p>a(x+y)² + b(x+y) + c = ax² + bx + c + ay² + by + c + xy</p><p>ax² + 2axy + ay² + bx + by + c = ax² + ay² + bx + by + 2c + xy</p><p><strong>Step 2:</strong> Compare coefficients of xy: 2a = 1, so <strong>a = 1/2</strong></p><p><strong>Step 3:</strong> The constant terms give: c = 2c, so <strong>c = 0</strong></p><p><strong>Step 4:</strong> Use constraint a + b + c = 3:</p><p>1/2 + b + 0 = 3 ⟹ <strong>b = 5/2</strong></p><p><strong>Step 5:</strong> Therefore f(x) = (1/2)x² + (5/2)x = (1/2)x(x+5)</p><p><strong>Step 6:</strong> Calculate the sum:</p><p>∑_{n=1}^{10} f(n) = ∑_{n=1}^{10} (1/2)n(n+5) = (1/2)∑_{n=1}^{10} (n² + 5n)</p><p>= (1/2)[∑n² + 5∑n] = (1/2)[10·11·21/6 + 5·10·11/2]</p><p>= (1/2)[385 + 275] = (1/2)(660) = <strong>330</strong></p><p>∴ Answer: B</p>
Correct Answer: B