Question

Let \( f:\mathbb{N}\to\mathbb{N} \) be such that \( f(1)=2 \) and \( f(x+y)=f(x)f(y) \). If \( \sum_{k=1}^{n} f(a+k)=16(2^n-1) \), then \( a \) is

• A. \( 3 \)
• B. \( 4 \)
• C. \( 5 \)
• D. \( 6 \)
• E. \( 7 \)

Hint:

Functional equations like \( f(x+y)=f(x)f(y) \) usually give exponential functions.

Answer 1

The Correct Option is B

Concept: Functional equation of exponential type.

Step 1: Since \( f(x+y)=f(x)f(y) \), assume: \[ f(n)=2^n \]

Step 2: Write sum: \[ \sum f(a+k)=\sum 2^{a+k} \]

Step 3: Factor: \[ =2^a \sum_{k=1}^n 2^k \]

Step 4: Use geometric sum: \[ =2^a (2^{n+1}-2) \]

Step 5: Compare: \[ 2^a(2^{n+1}-2)=16(2^n-1) \] Simplify: \[ 2^{a+1}(2^n-1)=16(2^n-1) \Rightarrow 2^{a+1}=16=2^4 \Rightarrow a=3 \]