Sets, Relations & Functions
Grade 11
Question:
<p>Let \(\sum_{k=1}^{10}f(a+k)=16(2^{10}-1)\), \(f(x+y)=f(x)f(y)\), \(f(1)=2\). Find \(a\).</p>
Step-by-Step Solution
<div class="solution"><p>f(n)=2ⁿ. Sum=2ᵃ·\Sigma2ᵏ(k=1..10)=2ᵃ·2(2^1^0-1)=16(2^1^0-1) \to 2ᵃ⁺^1=16=2^4 \to a=3.</p><p><strong>Answer: a=3</strong></p><div class="trap-box"><strong>Trap:</strong> f(x+y)=f(x)f(y) is multiplicative-on-addition, not f(xy).<div class="key-concept"><strong>Key Concept:</strong> Additive-input multiplicative f \to f(n)=f(1)ⁿ
Correct Answer: 3