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>165</p>
<p>190</p>
<p>255</p>
<p>330</p>
Step-by-Step Solution
Key Concept: Use the functional equation f(x+y) = f(x) + f(y) + xy to determine coefficients by substituting strategic values (x=y=1, x=1, y=0), then recognize that the resulting f(n) forms an arithmetic progression of second differences.
<p><strong>Step 1: Find coefficients using the functional equation</strong></p><p>Given: f(x) = ax² + bx + c with f(x+y) = f(x) + f(y) + xy</p><p>Substitute x = y = 1:</p><p>f(2) = f(1) + f(1) + 1</p><p>4a + 2b + c = 2(a + b + c) + 1</p><p>4a + 2b + c = 2(3) + 1 = 7</p><p>So: 4a + 2b + c = 7 ... (i)</p><p><strong>Step 2: Use another substitution</strong></p><p>Substitute y = 0:</p><p>f(x) = f(x) + f(0) + 0</p><p>This gives f(0) = 0, so c = 0</p><p><strong>Step 3: Apply the condition a + b + c = 3</strong></p><p>With c = 0: a + b = 3, so b = 3 - a</p><p>From equation (i): 4a + 2(3-a) + 0 = 7</p><p>4a + 6 - 2a = 7</p><p>2a = 1, so a = 1/2 and b = 5/2</p><p>Therefore: f(x) = ½x² + (5/2)x</p><p><strong>Step 4: Calculate f(n) for n = 1 to 10</strong></p><p>f(n) = ½n² + (5/2)n = (n² + 5n)/2 = n(n+5)/2</p><p><strong>Step 5: Compute the sum</strong></p><p>∑(n=1 to 10) f(n) = ½∑(n=1 to 10)(n² + 5n)</p><p>= ½[∑n² + 5∑n]</p><p>= ½[10(11)(21)/6 + 5·10(11)/2]</p><p>= ½[385 + 275]</p><p>= ½(660) = 330</p><p><strong>∴ Answer: 330</strong></p>
Correct Answer: D