Question

Consider a function ƒ satisfying ƒ(x + y) = ƒ(x)ƒ(y) where x,y are positive integers , and ƒ(1) = 2 . If ƒ(a+1) + ƒ(a+2) + ... + ƒ(a+n) = 16(2n - 1) then a is equal to 

• A. 4
• B. 1
• C. 3
• D. 2

Answer 1

The Correct Option is C

Given: 

\[ f(a+1) + f(a+2) + \cdots + f(a+n) = 16(2^n - 1) \]

Step 1: Substitution

Let us assume:

\[ f(a+1) + f(a+2) + \cdots + f(a+n) = f(a)f(1) + f(a)f(2) + \cdots + f(a)f(n) \Rightarrow f(a)(f(1) + f(2) + \cdots + f(n)) = 16(2^n - 1) \]

Step 2: Plug in \( n = 1 \)

\[ f(a)f(1) = 16(2^1 - 1) = 16 \Rightarrow f(a) \cdot f(1) = 16 \]

Assume \( f(1) = 2 \), then:

\[ f(a) = \frac{16}{2} = 8 \]

Step 3: Derive General Form

\[ f(1) + f(2) + \cdots + f(n) = \frac{16(2^n - 1)}{f(a)} = \frac{16(2^n - 1)}{8} = 2(2^n - 1) \]

Step 4: Compute Individual Values

• For \( n = 2 \):
  \[ f(1) + f(2) = 2(2^2 - 1) = 2(4 - 1) = 6 \Rightarrow f(2) = 6 - 2 = 4 \]
• For \( n = 3 \):
  \[ f(1) + f(2) + f(3) = 2(8 - 1) = 14 \Rightarrow f(3) = 14 - 2 - 4 = 8 \]

Conclusion:

Since \( f(3) = 8 = f(a) \Rightarrow a = 3 \)