Sets, Relations & Functions
Functional equations
Grade 11
Question:
<p>If \(f(x)\) satisfies the relation \(f(x + y) = f(x) + f(y)\) for all \(x, y \in \mathbb{R}\) and \(f(1) = 5\), then find \(\displaystyle\sum_{n=1}^{m} f(n)\). Also prove that \(f(x)\) is odd.</p>
Step-by-Step Solution
Key Concept: Cauchy's functional equation f(x+y)=f(x)+f(y) with f(1)=5 uniquely determines f(n)=5n for all integers. The sum becomes an arithmetic series: Σf(n) = 5Σn = 5m(m+1)/2. Oddness follows from f(0)=0 and f(-x)=-f(x).
<p><strong>Step 1 (Find f(n)):</strong> From f(x+y)=f(x)+f(y), set y=1 repeatedly:<br/>f(2)=f(1+1)=f(1)+f(1)=2f(1)=10<br/>f(3)=f(2+1)=f(2)+f(1)=10+5=15<br/>By induction: f(n)=nf(1)=5n for all n∈ℕ</p><p><strong>Step 2 (Prove f is odd):</strong> Setting x=y=0: f(0)=f(0)+f(0)⟹f(0)=0<br/>For any x∈ℝ: 0=f(0)=f(x+(-x))=f(x)+f(-x)<br/>Therefore: f(-x)=-f(x), so f is odd</p><p><strong>Step 3 (Compute the sum):</strong> <br/>∑ₙ₌₁ᵐ f(n) = ∑ₙ₌₁ᵐ 5n = 5∑ₙ₌₁ᵐ n = 5·m(m+1)/2</p><p><strong>∴ Answer: 5m(m+1)/2</strong><br/><em>Note: The functional equation characterizes f(x)=5x for all x∈ℝ (assuming continuity or monotonicity), making f both odd and linear.</em></p>
Correct Answer: 5m(m+1)/2