Question

If f(x) is a function such that f(x+y) = f(x)+ f(y) and f(1) = 7 then  \( \sum_{r=1}^{n}\) f(r) = 

• A. \(\frac{7n}{2}\)
• B. \(\frac{7(n+1)}{2}\)
• C. 7n(n+1)
• D. \(\frac{7n(n+1)}{2}\)

Answer 1

The Correct Option is D

To solve the problem, we are given a function \( f(x) \) such that:

• \( f(x + y) = f(x) + f(y) \)
• \( f(1) = 7 \)

1. Use the Property of the Function:
The functional equation \( f(x + y) = f(x) + f(y) \) is the definition of a linear function. Specifically, a function of the form \( f(x) = kx \) satisfies this equation for all real numbers.

2. Determine the Constant:
We are given \( f(1) = 7 \). If \( f(x) = kx \), then:
\[ f(1) = k \cdot 1 = 7 \Rightarrow k = 7 \] So, the function is \( f(x) = 7x \)

3. Evaluate the Summation:
We are to find: \[ \sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} 7r = 7 \sum_{r=1}^{n} r \] We know: \[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \] Therefore: \[ \sum_{r=1}^{n} f(r) = 7 \cdot \frac{n(n+1)}{2} = \frac{7n(n+1)}{2} \]

Final Answer:
The value of \( \sum_{r=1}^{n} f(r) \) is \( \frac{7n(n+1)}{2} \)