Question

Let \( f: \mathbb{R} \rightarrow \mathbb{R} \) be defined by \( f(x + y) = f(x) + 12y \), for all \( x, y \in \mathbb{R} \). If \( f(1) = 6 \), then the value of \( \sum_{r=1}^n f(r) \) is:

• A. \( n^2 \)
• B. \( 5n^2 \)
• C. \( 6n^2 \)
• D. \( \frac{3n(n + 1)}{2} \)

Hint:

When dealing with functions defined recursively with parameters, look for patterns or simplify the function using known values or conditions, such as \( f(1) \).

Answer 1

The Correct Option is C

Step 1: Identify the structure of the function.
- Given \( f(x + y) = f(x) + 12y \), we need to express \( f(x) \) in a more usable form.
- Set \( y = 0 \), then we get \( f(x) = f(x) + 12(0) \), which gives no new information, but it suggests a relationship where the function depends on both \( x \) and \( y \).
- Let’s assume a form for \( f(x) \) that will allow us to use the given condition \( f(1) = 6 \).
Step 2: Assume a quadratic form for \( f(x) \).
- Based on the structure \( f(x + y) = f(x) + 12y \), it suggests a quadratic function. Let \( f(x) = ax^2 + bx + c \).
- Now, substitute into the original equation: \[ f(x + y) = a(x + y)^2 + b(x + y) + c \] Expanding gives: \[ f(x + y) = a(x^2 + 2xy + y^2) + b(x + y) + c = ax^2 + 2axy + ay^2 + bx + by + c \] Comparing this with \( f(x) + 12y = ax^2 + bx + c + 12y \), we identify that \( 2axy = 12y \), leading to \( a = 6 \).
Step 3: Solve for constants.
- Using \( a = 6 \), we have \( f(x) = 6x^2 + bx - b \).
- From the given \( f(1) = 6 \), we substitute: \[ 6(1)^2 + b(1) - b = 6 \quad \Rightarrow \quad 6 + b - b = 6 \quad \Rightarrow \quad 6 = 6, \] which holds true. Thus, \( f(x) = 6x^2 \).
Step 4: Find the sum \( \sum_{r=1}^n f(r) \).
- We need to compute \( \sum_{r=1}^n f(r) = \sum_{r=1}^n (6r^2) \).
- The sum of squares of the first \( n \) integers is: \[ \sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}. \] - Therefore, \[ \sum_{r=1}^n 6r^2 = 6 \cdot \frac{n(n+1)(2n+1)}{6} = n(n+1)(2n+1). \]