If `""(n)C_(0), ""(n)C_(1), ""(n)C_(2), ...., ""(n)C_(n), ` denote the binomial coefficients in the expansion of `(1 + x)^(n) and p + q =1` ` sum_(r=0)^(n) r^(2 " "^n)C_(r) p^(r) q^(n-r) = ` .

Options

A. npq
B. np (p+q)
C. `np (np + q)`
D. ` np (p + nq)`

Explanation

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

\[ \sum_{r=0}^{n} r^2 \binom{n}{r} p^r q^{n-r} \]

where \( p + q = 1 \).

### Step-by-step Solution:

1. **Understanding the Summation**:
We start with the expression \(\sum_{r=0}^{n} r^2 \binom{n}{r} p^r q^{n-r}\). This summation involves binomial coefficients and powers of \(p\) and \(q\).

2. **Using the Identity for \(r \binom{n}{r}\)**:
We can use the identity:
\[ r \binom{n}{r} = n \binom{n-1}{r-1} \]
This allows us to rewrite the summation:
\[ \sum_{r=0}^{n} r^2 \binom{n}{r} p^r q^{n-r} = \sum_{r=0}^{n} r \cdot r \binom{n}{r} p^r q^{n-r} \]
We can express \(r\) as \(n \cdot \binom{n-1}{r-1}\), leading to:
\[ = n \sum_{r=1}^{n} r \binom{n-1}{r-1} p^r q^{n-r} \]

3. **Changing the Index of Summation**:
We change the index of summation by letting \(k = r - 1\):
\[ = n \sum_{k=0}^{n-1} (k+1) \binom{n-1}{k} p^{k+1} q^{(n-1)-k} \]
This can be split into two parts:
\[ = n \left( \sum_{k=0}^{n-1} \binom{n-1}{k} p^{k+1} q^{(n-1)-k} + \sum_{k=0}^{n-1} k \binom{n-1}{k} p^{k+1} q^{(n-1)-k} \right) \]

4. **Evaluating the First Summation**:
The first summation simplifies to:
\[ = n p \sum_{k=0}^{n-1} \binom{n-1}{k} p^k q^{(n-1)-k} = n p (p + q)^{n-1} = n p \]

5. **Evaluating the Second Summation**:
The second summation can be evaluated using the identity:
\[ k \binom{n-1}{k} = (n-1) \binom{n-2}{k-1} \]
Thus,
\[ \sum_{k=0}^{n-1} k \binom{n-1}{k} p^{k+1} q^{(n-1)-k} = (n-1) p \sum_{k=1}^{n-1} \binom{n-2}{k-1} p^{k-1} q^{(n-2)-(k-1)} \]
This simplifies to:
\[ = (n-1) p (p + q)^{n-2} = (n-1) p \]

6. **Combining Results**:
Combining both parts, we have:
\[ \sum_{r=0}^{n} r^2 \binom{n}{r} p^r q^{n-r} = n p + (n-1) p = n^2 p \]

7. **Final Result**:
Since \(p + q = 1\), the final result is:
\[ \sum_{r=0}^{n} r^2 \binom{n}{r} p^r q^{n-r} = n^2 p \]