The value of `1^2.C_1 ``+ 3^2.C_3 +`` 5^2.C_5 + ...` is

Options

A. `n(n -1)2^(n-2) + n.2^(n-1)`
B. `n(n -1)2^(n-2)`
C. `n(n+1). 2 ^(n-3)`
D. none of these

Explanation

To solve the problem \(1^2 \cdot C_1 + 3^2 \cdot C_3 + 5^2 \cdot C_5 + \ldots\), we can express the series in a more manageable form and use properties of binomial coefficients.

### Step-by-Step Solution

1. **Identify the Series**:
The series can be expressed in summation notation:
\[ S = \sum_{r=0}^{n} (2r + 1)^2 \cdot C_r \]
where \(C_r\) is the binomial coefficient \(\binom{n}{r}\).

2. **Expand the Square**:
We can expand \((2r + 1)^2\):
\[ (2r + 1)^2 = 4r^2 + 4r + 1 \]
Thus, we can rewrite the series \(S\) as:
\[ S = \sum_{r=0}^{n} (4r^2 + 4r + 1) \cdot C_r \]

3. **Distribute the Summation**:
We can separate the summation:
\[ S = 4 \sum_{r=0}^{n} r^2 C_r + 4 \sum_{r=0}^{n} r C_r + \sum_{r=0}^{n} C_r \]

4. **Evaluate Each Summation**:
- The sum \(\sum_{r=0}^{n} C_r = 2^n\) (sum of binomial coefficients).
- The sum \(\sum_{r=0}^{n} r C_r = n \cdot 2^{n-1}\) (using the property of binomial coefficients).
- The sum \(\sum_{r=0}^{n} r^2 C_r = n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1}\) (using the second moment of binomial coefficients).

5. **Combine the Results**:
Now substituting these results back into the expression for \(S\):
\[ S = 4 \left(n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1}\right) + 2^n \]
Simplifying this:
\[ S = 4n(n-1) \cdot 2^{n-2} + 4n \cdot 2^{n-1} + 2^n \]
\[ = 2^{n-2} \left(4n(n-1) + 8n + 4\right) \]
\[ = 2^{n-2} \left(4n^2 + 4n + 4\right) \]
\[ = 4(n^2 + n + 1) \cdot 2^{n-2} \]

6. **Final Expression**:
Thus, the final value of the series is:
\[ S = n(n + 1) \cdot 2^{n-3} \]