If `(1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + ... + C_(n) x^(n)`, then value of `C_(0)^(2) + 2C_(1)^(2) + 3C_(2)^(2) + ... + (n + 1) C^(2)n` is

Options

A. `(2n+1)(""^(2n)C_(n))`
B. `(2n-1)(""^(2n)C_(n))`
C. `(n/2+1)(""^(2n)C_(n))`
D. `(1+n/2)(""^(2n-1)C_(n))`

Answer

To find the value of \( C_0^2 + 2C_1^2 + 3C_2^2 + \ldots + (n+1)C_n^2 \), we will use the properties of binomial coefficients and some algebraic manipulation.

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial expansion of \( (1 + x)^n \) is given by:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]
where \( C_k = \binom{n}{k} \) for \( k = 0, 1, 2, \ldots, n \).

2. **Setting Up the Summation**:
We need to evaluate:
\[ S = C_0^2 + 2C_1^2 + 3C_2^2 + \ldots + (n+1)C_n^2 \]
This can be rewritten using summation notation:
\[ S = \sum_{k=0}^{n} (k+1) C_k^2 \]

3. **Using the Identity for Binomial Coefficients**:
We can use the identity:
\[ \sum_{k=0}^{n} k C_k^2 = n C_n \]
This gives us a way to separate the summation into two parts:
\[ S = \sum_{k=0}^{n} C_k^2 + \sum_{k=0}^{n} k C_k^2 \]

4. **Calculating \( \sum_{k=0}^{n} C_k^2 \)**:
There is a known result that states:
\[ \sum_{k=0}^{n} C_k^2 = C_{2n}^n \]

5. **Calculating \( \sum_{k=0}^{n} k C_k^2 \)**:
Using the identity mentioned earlier:
\[ \sum_{k=0}^{n} k C_k^2 = n C_n \]

6. **Combining the Results**:
Now substituting back into our expression for \( S \):
\[ S = C_{2n}^n + n C_n \]

7. **Final Expression**:
Thus, the value of \( C_0^2 + 2C_1^2 + 3C_2^2 + \ldots + (n+1)C_n^2 \) is:
\[ S = C_{2n}^n + n C_n \]