Question

If \( f(1)=1 \), \( f(2n)=f(n) \) and \( f(2n+1)=(f(n))^2 - 2 \) for \( n=1,2,3,\ldots \), then the value of \( f(1)+f(2)+\cdots+f(25) \) is equal to

• A. \(1\)
• B. \(-15\)
• C. \(-17\)
• D. \(-1\)
• E. \(13\)

Hint:

In recursive functions involving even and odd cases, repeatedly reduce the argument. Numbers that reduce to the base case directly (like powers of 2 here) behave differently from others—identify them to simplify counting.

Answer 1

The Correct Option is B

Concept: The function is defined recursively depending on whether the input is even or odd:
• If \(n\) is even: \( f(2n) = f(n) \), which reduces the argument.
• If \(n\) is odd: \( f(2n+1) = (f(n))^2 - 2 \). Thus, every value of \(f(n)\) can be reduced step-by-step to smaller values until it ultimately depends on \(f(1)\).

Step 1: Base value
\[ f(1) = 1 \]

Step 2: Understanding the recursive behavior
Let us observe what happens when values are substituted:
• If \( f(n) = 1 \), then \[ f(2n+1) = 1^2 - 2 = -1 \]
• If \( f(n) = -1 \), then \[ f(2n+1) = (-1)^2 - 2 = 1 - 2 = -1 \] Thus, once the value becomes \(-1\), it remains \(-1\) for all further odd expansions.

Step 3: Compute initial values explicitly
\[ \begin{aligned} f(1)&=1 \\ f(2)&=f(1)=1 \\ f(3)&=(f(1))^2-2= -1 \\ f(4)&=f(2)=1 \\ f(5)&=(f(2))^2-2= -1 \\ f(6)&=f(3)= -1 \\ f(7)&=(f(3))^2-2= -1 \\ f(8)&=f(4)=1 \\ f(9)&=(f(4))^2-2= -1 \\ f(10)&=f(5)= -1 \end{aligned} \] Continuing similarly, we see a clear pattern.

Step 4: Identify which numbers give \( f(n)=1 \)
A number gives \(f(n)=1\) only if repeated division by 2 eventually reduces it to 1. Thus, numbers of the form \(2^k\) give: \[ f(1)=1,\quad f(2)=1,\quad f(4)=1,\quad f(8)=1,\quad f(16)=1 \] So, within 1 to 25: \[ \text{Numbers giving } f(n)=1: \{1,2,4,8,16\} \] Total count: \[ 5 \]

Step 5: Count remaining values
Total numbers from 1 to 25: \[ 25 \] Thus, \[ \text{Numbers giving } f(n)=-1 = 25 - 5 = 20 \]

Step 6: Compute the required sum
\[ \sum_{n=1}^{25} f(n) = 5(1) + 20(-1) \] \[ = 5 - 20 = -15 \]

Step 7: Final Answer
\[ \boxed{-15} \]