If `x+y=1`, then `sum_(r=0)^(n)r^(2)""^(n)C_(r )x^(r )y^(n-r)` equals.

Options

A. nxy
B. nx (x + yn)
C. nx(nx+y)
D. none of these

Answer

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

\[ \sum_{r=0}^{n} r^2 \binom{n}{r} x^r y^{n-r} \]

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

### Step 1: Rewrite \( r^2 \)

We can express \( r^2 \) as:

\[ r^2 = r(r-1) + r \]

This allows us to split the original sum into two separate sums:

\[ \sum_{r=0}^{n} r^2 \binom{n}{r} x^r y^{n-r} = \sum_{r=0}^{n} r(r-1) \binom{n}{r} x^r y^{n-r} + \sum_{r=0}^{n} r \binom{n}{r} x^r y^{n-r} \]

### Step 2: Evaluate the first sum

The first sum can be simplified:

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

Using the identity \( r(r-1) \binom{n}{r} = n(n-1) \binom{n-2}{r-2} \), we can rewrite the sum:

\[ \sum_{r=2}^{n} n(n-1) \binom{n-2}{r-2} x^r y^{n-r} \]

Changing the index of summation by letting \( k = r - 2 \):

\[ = n(n-1) x^2 \sum_{k=0}^{n-2} \binom{n-2}{k} x^k y^{(n-2)-k} \]

By the Binomial Theorem, this sum equals \( (x + y)^{n-2} = 1^{n-2} = 1 \).

Thus, we have:

\[ \sum_{r=0}^{n} r(r-1) \binom{n}{r} x^r y^{n-r} = n(n-1) x^2 \]

### Step 3: Evaluate the second sum

The second sum is:

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

Using the identity \( r \binom{n}{r} = n \binom{n-1}{r-1} \), we can rewrite it as:

\[ n \sum_{r=1}^{n} \binom{n-1}{r-1} x^r y^{n-r} \]

Changing the index of summation by letting \( k = r - 1 \):

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

Again, by the Binomial Theorem, this sum equals \( (x + y)^{n-1} = 1^{n-1} = 1 \).

Thus, we have:

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

### Step 4: Combine the results

Now, we can combine the results of the two sums:

\[ \sum_{r=0}^{n} r^2 \binom{n}{r} x^r y^{n-r} = n(n-1)x^2 + nx \]

Factoring out \( nx \):

\[ = nx \left( (n-1)x + 1 \right) = nx(nx + y) \]

Since \( y = 1 - x \):

\[ = nx(1) = nx \]

### Final Result

Thus, the final result is:

\[ \sum_{r=0}^{n} r^2 \binom{n}{r} x^r y^{n-r} = nxy \]