`f(1)=1 and f(n)=2sum_(r=1)^(n-1) f (r)`. Then `sum_(n=1)^mf(n)` is equal to
(A)`3^m-1`
(B)`3^m`
(C)`3^(m-1)`
(D)none of these

Options

A. 3 ^(m) -1
B. `3 ^(m)`
C. `3 ^(m-1)`
D. `none of these`

Answer

To solve the problem, we need to find the function \( f(n) \) based on the given recurrence relation and then compute the sum \( \sum_{n=1}^{m} f(n) \).

### Step 1: Calculate \( f(2) \)
Given:
- \( f(1) = 1 \)
- \( f(n) = 2 \sum_{r=1}^{n-1} f(r) \)

For \( n = 2 \):
\[ f(2) = 2 \sum_{r=1}^{1} f(r) = 2 f(1) = 2 \times 1 = 2 \]

### Step 2: Calculate \( f(3) \)
For \( n = 3 \):
\[ f(3) = 2 \sum_{r=1}^{2} f(r) = 2 (f(1) + f(2)) = 2 (1 + 2) = 2 \times 3 = 6 \]

### Step 3: Calculate \( f(4) \)
For \( n = 4 \):
\[ f(4) = 2 \sum_{r=1}^{3} f(r) = 2 (f(1) + f(2) + f(3)) = 2 (1 + 2 + 6) = 2 \times 9 = 18 \]

### Step 4: Identify a pattern
From our calculations:
- \( f(1) = 1 = 2 \cdot 3^0 \)
- \( f(2) = 2 = 2 \cdot 3^0 \)
- \( f(3) = 6 = 2 \cdot 3^1 \)
- \( f(4) = 18 = 2 \cdot 3^2 \)

We can see a pattern forming:
\[ f(n) = 2 \cdot 3^{n-2} \quad \text{for } n \geq 2 \]

### Step 5: General formula for \( f(n) \)
Thus, we can write:
\[ f(n) = \begin{cases} 1 & \text{if } n = 1 \\ 2 \cdot 3^{n-2} & \text{if } n \geq 2 \end{cases} \]

### Step 6: Calculate \( \sum_{n=1}^{m} f(n) \)
Now we need to compute:
\[ \sum_{n=1}^{m} f(n) = f(1) + \sum_{n=2}^{m} f(n) \]
\[ = 1 + \sum_{n=2}^{m} 2 \cdot 3^{n-2} \]

### Step 7: Simplify the summation
The summation \( \sum_{n=2}^{m} 2 \cdot 3^{n-2} \) can be rewritten:
\[ = 2 \sum_{k=0}^{m-2} 3^k \quad \text{(where \( k = n-2 \))} \]

This is a geometric series with:
- First term \( a = 1 \) (when \( k = 0 \))
- Common ratio \( r = 3 \)
- Number of terms \( m-1 \)

The sum of a geometric series is given by:
\[ S = \frac{a(r^n - 1)}{r - 1} \]
Substituting the values:
\[ S = \frac{1(3^{m-1} - 1)}{3 - 1} = \frac{3^{m-1} - 1}{2} \]

Thus,
\[ \sum_{n=2}^{m} f(n) = 2 \cdot \frac{3^{m-1} - 1}{2} = 3^{m-1} - 1 \]

### Final Calculation
Now, substituting back:
\[ \sum_{n=1}^{m} f(n) = 1 + (3^{m-1} - 1) = 3^{m-1} \]

### Conclusion
The final answer is:
\[ \sum_{n=1}^{m} f(n) = 3^{m-1} \]

### Answer
(C) \( 3^{m-1} \)