Relations & Functions
Functional Equations
Grade 12
Question:
<p>Find the natural number \(a\) for which \(\displaystyle\sum_{k=1}^{n} f(a+k) = 16(2^n - 1)\), where the function \(f\) satisfies \(f(x+y) = f(x)f(y)\) for all natural numbers \(x, y\) and \(f(1) = 2\).</p>
Step-by-Step Solution
Key Concept: The functional equation f(x+y) = f(x)f(y) with f(1) = 2 identifies f as an exponential function f(x) = 2^x. The sum then becomes a geometric series that can be matched to the given form to solve for a.
<p><strong>Step 1:</strong> Determine the function f(x).</p><p>Given: f(x+y) = f(x)f(y) and f(1) = 2</p><p>This is Cauchy's exponential functional equation. For natural numbers, f(n) = f(1)^n = 2^n</p><p>We can verify: f(x+y) = 2^(x+y) = 2^x · 2^y = f(x)f(y) ✓</p><p><strong>Step 2:</strong> Express the sum using f(x) = 2^x.</p><p>∑_{k=1}^{n} f(a+k) = ∑_{k=1}^{n} 2^(a+k) = ∑_{k=1}^{n} 2^a · 2^k = 2^a ∑_{k=1}^{n} 2^k</p><p><strong>Step 3:</strong> Evaluate the geometric series.</p><p>∑_{k=1}^{n} 2^k = 2 + 2² + 2³ + ... + 2^n = 2(2^n - 1)/(2-1) = 2(2^n - 1)</p><p><strong>Step 4:</strong> Match with the given expression.</p><p>2^a · 2(2^n - 1) = 16(2^n - 1)</p><p>2^a · 2 = 16</p><p>2^(a+1) = 2^4</p><p>a + 1 = 4</p><p>∴ Answer: a = 3</p>
Correct Answer: 3