If x+y= 1, then value of `sum_(r=0)^(n)(r)(""^(n)C_(r))x^(n-r)y^(r)` is

Options

A. 1
B. 0
C. nx
D. ny

Answer

To solve the problem, we need to evaluate the expression:

\[ \sum_{r=0}^{n} r \binom{n}{r} x^{n-r} y^r \]

given that \( x + y = 1 \).

### Step-by-Step Solution:

1. **Understanding the Summation**:
The expression involves a summation where \( r \) is multiplied by the binomial coefficient \( \binom{n}{r} \) and the powers of \( x \) and \( y \).

2. **Using the Property of Binomial Coefficients**:
We know that:
\[ r \binom{n}{r} = n \binom{n-1}{r-1} \]
This allows us to rewrite the summation:
\[ \sum_{r=0}^{n} r \binom{n}{r} x^{n-r} y^r = \sum_{r=1}^{n} n \binom{n-1}{r-1} x^{n-r} y^r \]

3. **Changing the Index of Summation**:
By changing the index of summation (let \( k = r - 1 \)), we can rewrite the summation:
\[ = n \sum_{k=0}^{n-1} \binom{n-1}{k} x^{n-1-k} y^{k+1} \]
This can be further simplified to:
\[ = n y \sum_{k=0}^{n-1} \binom{n-1}{k} x^{n-1-k} y^k \]

4. **Recognizing the Binomial Theorem**:
The summation \( \sum_{k=0}^{n-1} \binom{n-1}{k} x^{n-1-k} y^k \) is the expansion of \( (x + y)^{n-1} \):
\[ = (x + y)^{n-1} \]

5. **Substituting \( x + y = 1 \)**:
Since \( x + y = 1 \), we have:
\[ (x + y)^{n-1} = 1^{n-1} = 1 \]

6. **Final Expression**:
Putting it all together, we get:
\[ \sum_{r=0}^{n} r \binom{n}{r} x^{n-r} y^r = n y \cdot 1 = n y \]

### Conclusion:
Thus, the value of the summation is:
\[ \sum_{r=0}^{n} r \binom{n}{r} x^{n-r} y^r = n y \]