If f is a real valued function such that `f(x+y)=f(x)+f(y) and f(1)=5`, then the value of f(100) is

Options

A. 200
B. 300
C. 400
D. 500

Answer

To solve the problem, we start with the functional equation given:

1. **Given:** \( f(x+y) = f(x) + f(y) \)
2. **Also given:** \( f(1) = 5 \)

We need to find \( f(100) \).

### Step 1: Find \( f(2) \)

Using the functional equation, let \( x = 1 \) and \( y = 1 \):

\[ f(1 + 1) = f(1) + f(1) \implies f(2) = 2f(1) \]

Substituting \( f(1) = 5 \):

\[ f(2) = 2 \times 5 = 10 \]

### Step 2: Find \( f(3) \)

Now, let \( x = 2 \) and \( y = 1 \):

\[ f(2 + 1) = f(2) + f(1) \implies f(3) = f(2) + f(1) \]

Substituting the values we have:

\[ f(3) = 10 + 5 = 15 \]

### Step 3: Find \( f(4) \)

Next, let \( x = 2 \) and \( y = 2 \):

\[ f(2 + 2) = f(2) + f(2) \implies f(4) = 2f(2) \]

Substituting \( f(2) = 10 \):

\[ f(4) = 2 \times 10 = 20 \]

### Step 4: Generalize \( f(n) \)

From the calculations above, we can see a pattern emerging. It appears that:

\[ f(n) = n \times f(1) \]

To prove this by induction, we assume it holds for \( n \) and \( n-1 \):

- Base case: For \( n = 1 \), \( f(1) = 1 \times 5 = 5 \) (true).
- Inductive step: Assume \( f(k) = k \times 5 \) holds for \( k \). Then for \( k+1 \):

\[ f(k+1) = f(k) + f(1) = (k \times 5) + 5 = (k + 1) \times 5 \]

Thus, by induction, \( f(n) = n \times 5 \) holds for all natural numbers \( n \).

### Step 5: Find \( f(100) \)

Now we can find \( f(100) \):

\[ f(100) = 100 \times f(1) = 100 \times 5 = 500 \]

### Final Answer

Thus, the value of \( f(100) \) is \( \boxed{500} \).

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