If `f(x) = ""^(40)C_(1).x(1-x)^(39) + 2.""^(40)C_(2).x^(2)(1-x)^(38)+3.""^(40)C_(3).x^(3)(1-x)^(37)+``"….."+40.""^(40)C_(40).x^(40)`, then the value of `f(3)` is

Options

A. a. 120
B. b. 150
C. c. 200
D. d. 240

Explanation

To solve the problem, we need to find the value of the function \( f(x) \) defined as:

\[ f(x) = \sum_{r=1}^{40} r \cdot \binom{40}{r} x^r (1-x)^{40-r} \]

### Step 1: Write the general term
The general term \( T_r \) of the series can be expressed as:
\[ T_r = r \cdot \binom{40}{r} x^r (1-x)^{40-r} \]

### Step 2: Simplify the general term
Using the property of binomial coefficients, we can rewrite \( T_r \):
\[ T_r = r \cdot \frac{40!}{r!(40-r)!} x^r (1-x)^{40-r} \]
This simplifies to:
\[ T_r = \frac{40!}{(r-1)!(40-r)!} x^r (1-x)^{40-r} = 40 \cdot \binom{39}{r-1} x^r (1-x)^{40-r} \]

### Step 3: Factor out constants
Now, we can factor out the constant \( 40 \) from the summation:
\[ f(x) = 40 \sum_{r=1}^{40} \binom{39}{r-1} x^r (1-x)^{40-r} \]

### Step 4: Change the index of summation
Let \( k = r - 1 \). Then, when \( r = 1 \), \( k = 0 \) and when \( r = 40 \), \( k = 39 \). Thus, we can rewrite the summation:
\[ f(x) = 40 \sum_{k=0}^{39} \binom{39}{k} x^{k+1} (1-x)^{39-k} \]

### Step 5: Factor out \( x \)
We can factor out \( x \) from the summation:
\[ f(x) = 40x \sum_{k=0}^{39} \binom{39}{k} x^k (1-x)^{39-k} \]

### Step 6: Apply the binomial theorem
Using the binomial theorem, we know that:
\[ \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} = (a + b)^n \]
In our case, \( a = x \) and \( b = 1-x \), so:
\[ \sum_{k=0}^{39} \binom{39}{k} x^k (1-x)^{39-k} = (x + (1-x))^{39} = 1^{39} = 1 \]

### Step 7: Final expression for \( f(x) \)
Thus, we have:
\[ f(x) = 40x \cdot 1 = 40x \]

### Step 8: Calculate \( f(3) \)
Now, we need to find \( f(3) \):
\[ f(3) = 40 \cdot 3 = 120 \]

### Conclusion
The value of \( f(3) \) is \( 120 \).