`If a + b = k, ` when `a, b gt o` and
`S(k, n) = sum _(r=0)^(n) r^(2) (""^(n) C_(r) ) a^(r) cdot b^(n-r) ` , then

Options

A. `S(1,3) =3 (3a^(2)+ab) `
B. `S(2,4)=16(4a^(2)+ab)`
C. `S(3,5) = 25 (5a^(2) + ab) `
D. `S(4,6)=36(6a^(2)+ab)`

Explanation

To solve the given problem, we need to evaluate the expression \( S(k, n) = \sum_{r=0}^{n} r^2 \binom{n}{r} a^r b^{n-r} \) under the condition \( a + b = k \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ S(k, n) = \sum_{r=0}^{n} r^2 \binom{n}{r} a^r b^{n-r} \]
Here, \( \binom{n}{r} \) is the binomial coefficient, and \( a^r b^{n-r} \) represents the terms in the binomial expansion.

2. **Using the Identity for \( r^2 \)**:
We can express \( r^2 \) in terms of binomial coefficients:
\[ r^2 = r \cdot r = r \cdot (n \cdot \binom{n-1}{r-1}) = n \cdot \binom{n-1}{r-1} + r(n-1) \cdot \binom{n-2}{r-2} \]
Thus, we can rewrite \( S(k, n) \):
\[ S(k, n) = \sum_{r=0}^{n} \left( n \cdot \binom{n-1}{r-1} + r(n-1) \cdot \binom{n-2}{r-2} \right) a^r b^{n-r} \]

3. **Separating the Summation**:
We can separate the summation into two parts:
\[ S(k, n) = n \sum_{r=1}^{n} \binom{n-1}{r-1} a^r b^{n-r} + (n-1) \sum_{r=2}^{n} \binom{n-2}{r-2} a^r b^{n-r} \]

4. **Evaluating the First Summation**:
The first summation can be simplified using the binomial theorem:
\[ \sum_{r=1}^{n} \binom{n-1}{r-1} a^r b^{n-r} = a \sum_{r=0}^{n-1} \binom{n-1}{r} a^r b^{n-1-r} = a (a + b)^{n-1} = a k^{n-1} \]

5. **Evaluating the Second Summation**:
The second summation can also be simplified:
\[ (n-1) \sum_{r=2}^{n} \binom{n-2}{r-2} a^r b^{n-r} = (n-1) a^2 \sum_{r=0}^{n-2} \binom{n-2}{r} a^r b^{n-2-r} = (n-1) a^2 (a + b)^{n-2} = (n-1) a^2 k^{n-2} \]

6. **Combining the Results**:
Now, we combine both parts:
\[ S(k, n) = n a k^{n-1} + (n-1) a^2 k^{n-2} \]

7. **Final Expression**:
The final expression for \( S(k, n) \) is:
\[ S(k, n) = n a k^{n-1} + (n-1) a^2 k^{n-2} \]