If `C_(r)` be the coefficients of `x^(r)` in `(1 + x)^(n) `, then the value
of `sum_(r=0)^(n) (r + 1)^(2) C_(r) `, is

Options

A. `(n +1) (n +4) 2^(n-2)`
B. `( n+1)(n +4)2^(n-1)`
C. `( n+ 1)^(2) 2^(n-2)`
D. `(n +4)^(2) 2^(n-2)`

Explanation

To solve the problem, we need to find the value of the summation:

\[ S_n = \sum_{r=0}^{n} (r + 1)^2 C_r \]

where \( C_r \) is the coefficient of \( x^r \) in the expansion of \( (1 + x)^n \), which is given by \( C_r = \binom{n}{r} \).

### Step-by-Step Solution:

1. **Expand the term \((r + 1)^2\)**:
\[ (r + 1)^2 = r^2 + 2r + 1 \]
Therefore, we can rewrite the summation as:
\[ S_n = \sum_{r=0}^{n} (r^2 + 2r + 1) C_r = \sum_{r=0}^{n} r^2 C_r + 2 \sum_{r=0}^{n} r C_r + \sum_{r=0}^{n} C_r \]

2. **Evaluate \(\sum_{r=0}^{n} C_r\)**:
By the binomial theorem, we know:
\[ \sum_{r=0}^{n} C_r = (1 + 1)^n = 2^n \]

3. **Evaluate \(\sum_{r=0}^{n} r C_r\)**:
We can express \( r C_r \) as \( n C_{r-1} \):
\[ \sum_{r=0}^{n} r C_r = n \sum_{r=1}^{n} C_{r-1} = n \sum_{s=0}^{n-1} C_s = n \cdot 2^{n-1} \]

4. **Evaluate \(\sum_{r=0}^{n} r^2 C_r\)**:
We use the identity:
\[ \sum_{r=0}^{n} r^2 C_r = n(n-1) \sum_{r=2}^{n} C_{r-2} + n \sum_{r=1}^{n} C_{r-1} \]
This simplifies to:
\[ \sum_{r=0}^{n} r^2 C_r = n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1} \]

5. **Combine all parts**:
Now substituting back into \( S_n \):
\[ S_n = \left( n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1} \right) + 2(n \cdot 2^{n-1}) + 2^n \]
Simplifying this:
\[ S_n = n(n-1) \cdot 2^{n-2} + 3n \cdot 2^{n-1} + 2^n \]

6. **Factor out \( 2^{n-2} \)**:
\[ S_n = 2^{n-2} \left( n(n-1) + 6n + 4 \right) = 2^{n-2} (n^2 + 5n + 4) \]

7. **Final Result**:
The final answer can be factored:
\[ S_n = 2^{n-2} (n + 4)(n + 1) \]

### Conclusion:
Thus, the value of the summation \( S_n \) is:

\[ S_n = 2^{n-2} (n + 4)(n + 1) \]