Question

If \( f:\mathbb{R} \to \mathbb{R} \) is defined by \( f(x+y) = f(x) + f(y) \ \forall x, y \in \mathbb{R} \) and \( f(1) = 7 \), then \(\sum\limits_{r=1}^n f(r) =\)

• A. \( \frac{3n(n+2)}{4} \)
• B. \( \frac{n(n-1)}{2} \)
• C. \( \frac{7n(n+1)}{2} \)
• D. \( \frac{(n+1)(n+2)}{4} \)

Hint:

For additive functions satisfying \( f(x+y) = f(x) + f(y) \), try substituting \( f(x) = kx \) and use the given value to find \( k \).

Answer 1

The Correct Option is C

Step 1: Understand the functional equation.
Given \( f(x+y) = f(x) + f(y) \), this is Cauchy's functional equation. For such functions (under regularity conditions like continuity), the general solution is: \[ f(x) = kx \] where \( k \) is a constant. Step 2: Use initial condition.
Given \( f(1) = 7 \Rightarrow f(x) = 7x \) Step 3: Compute the sum.
\[ \sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} 7r = 7 \sum_{r=1}^{n} r = 7 \cdot \frac{n(n+1)}{2} \] \[ \boxed{ \sum_{r=1}^{n} f(r) = \frac{7n(n+1)}{2} } \]