If f (x) is a function such that f(x+y)=f(x)+f(y) and `f(1)=7" then "underset(r=1)overset(n)sum f(r)` is equal to :

Options

A. 7n/2
B. 7(n+1)2
C. 7n(n+1)
D. 7n(n+1)/2

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the functional equation
We are given that \( f(x+y) = f(x) + f(y) \). This is a well-known property of linear functions, suggesting that \( f(x) \) could be of the form \( f(x) = kx \) for some constant \( k \).

### Step 2: Use the given value
We also know that \( f(1) = 7 \). If we assume \( f(x) = kx \), then substituting \( x = 1 \) gives us:
\[ f(1) = k \cdot 1 = k \]
Thus, we have \( k = 7 \). Therefore, we can express the function as:
\[ f(x) = 7x \]

### Step 3: Calculate \( f(r) \) for \( r = 1, 2, \ldots, n \)
Now we need to find the summation \( \sum_{r=1}^{n} f(r) \):
\[ f(r) = 7r \]
So,
\[ \sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} 7r = 7 \sum_{r=1}^{n} r \]

### Step 4: Use the formula for the sum of the first \( n \) natural numbers
The sum of the first \( n \) natural numbers is given by the formula:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
Substituting this into our equation gives:
\[ \sum_{r=1}^{n} f(r) = 7 \cdot \frac{n(n+1)}{2} \]

### Step 5: Final expression
Thus, we find that:
\[ \sum_{r=1}^{n} f(r) = \frac{7n(n+1)}{2} \]

### Conclusion
The final answer is:
\[ \sum_{r=1}^{n} f(r) = \frac{7n(n+1)}{2} \]

---