If f(x) satisfies the relation, f(x+y)=f(x)+f(y) for all x,y `in` R and f(1)=5, then find `sum_(n=1)^(m)f(n)`. Also, prove that f(x) is odd.

Answer

To solve the problem step by step, we will follow the given functional equation and the information provided.

### Step 1: Establish the functional equation
We are given that:
\[ f(x + y) = f(x) + f(y) \]
for all \( x, y \in \mathbb{R} \).

### Step 2: Find \( f(0) \)
Let's substitute \( x = 0 \) and \( y = 0 \) into the functional equation:
\[ f(0 + 0) = f(0) + f(0) \implies f(0) = 2f(0) \]
This simplifies to:
\[ f(0) - 2f(0) = 0 \implies -f(0) = 0 \implies f(0) = 0 \]

### Step 3: Prove that \( f(x) \) is odd
Now, let’s substitute \( y = -x \):
\[ f(x + (-x)) = f(x) + f(-x) \implies f(0) = f(x) + f(-x) \]
Since we found that \( f(0) = 0 \), we have:
\[ 0 = f(x) + f(-x) \implies f(-x) = -f(x) \]
This shows that \( f(x) \) is an odd function.

### Step 4: Find \( f(n) \) for natural numbers
Given \( f(1) = 5 \), we can find \( f(2) \):
\[ f(2) = f(1 + 1) = f(1) + f(1) = 5 + 5 = 10 \]
Next, we find \( f(3) \):
\[ f(3) = f(1 + 2) = f(1) + f(2) = 5 + 10 = 15 \]
Now, for \( f(4) \):
\[ f(4) = f(2 + 2) = f(2) + f(2) = 10 + 10 = 20 \]
From this pattern, we can deduce that:
\[ f(n) = 5n \]
for \( n = 1, 2, 3, 4, \ldots \).

### Step 5: Find the summation \( \sum_{n=1}^{m} f(n) \)
Now we need to find:
\[ \sum_{n=1}^{m} f(n) = \sum_{n=1}^{m} 5n = 5 \sum_{n=1}^{m} n \]
Using the formula for the sum of the first \( m \) natural numbers:
\[ \sum_{n=1}^{m} n = \frac{m(m + 1)}{2} \]
Thus, we have:
\[ \sum_{n=1}^{m} f(n) = 5 \cdot \frac{m(m + 1)}{2} = \frac{5m(m + 1)}{2} \]

### Final Answer
The final result for the summation is:
\[ \sum_{n=1}^{m} f(n) = \frac{5m(m + 1)}{2} \]