The value of
`""(n)C_(1). X(1 - x )^(n-1) + 2 . ""^(n)C_(2) x^(2) (1 - x)^(n-2)`
` + 3. ""^(n)C_(3) x^(3) (1 - x)^(n-3) + ….+ n ""^(n)C_(n) x^(n) , n in ` N is

Options

A. nx
B. ` n(n -x)`
C. ` n (x-1)`
D. none of these

Explanation

To solve the given expression:

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

we will follow these steps:

### Step 1: Rewrite the expression
We can express \( r \cdot \binom{n}{r} \) in a different form. Recall that \( r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1} \). Thus, we can rewrite \( S_n \) as:

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

### Step 2: Factor out the constant
Since \( n \) is a constant with respect to the summation, we can factor it out:

\[ S_n = n \sum_{r=1}^{n} \binom{n-1}{r-1} x^r (1-x)^{n-r} \]

### Step 3: Change the index of summation
Next, we can change the index of summation by letting \( k = r - 1 \). When \( r = 1 \), \( k = 0 \) and when \( r = n \), \( k = n - 1 \). Thus, we have:

\[ S_n = n \sum_{k=0}^{n-1} \binom{n-1}{k} x^{k+1} (1-x)^{(n-1)-(k+1)} \]

This simplifies to:

\[ S_n = n x \sum_{k=0}^{n-1} \binom{n-1}{k} x^k (1-x)^{n-1-k} \]

### Step 4: Recognize the binomial expansion
The sum \( \sum_{k=0}^{n-1} \binom{n-1}{k} x^k (1-x)^{(n-1)-k} \) is the binomial expansion of \( (x + (1-x))^{n-1} = 1^{n-1} = 1 \).

### Step 5: Final result
Thus, we have:

\[ S_n = n x \cdot 1 = n x \]

### Conclusion
The value of the original expression is:

\[ \boxed{n x} \]