If `C_(0), C_(1), C_(2), …, C_(n)` denote the binomial coefficients
in the expansion of `(1 + x)^(n)`, then
`C_(0)""^(2) + 2 C_(1)""^(2) + 3C_(2)""^(2) + ...+ (n +1)C_(n)""^(2) =`

Options

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

Explanation

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

\[ C_0^2 + 2C_1^2 + 3C_2^2 + \ldots + (n + 1)C_n^2 \]

where \( C_k \) are the binomial coefficients from the expansion of \( (1 + x)^n \).

### Step-by-Step Solution:

**Step 1: Define the sum.**
Let \( S = C_0^2 + 2C_1^2 + 3C_2^2 + \ldots + (n + 1)C_n^2 \).

**Hint 1:** Recognize that \( S \) is a weighted sum of the squares of the binomial coefficients.

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**Step 2: Reverse the order of summation.**
We can express \( S \) in terms of \( C_{n-k} \) using the property of binomial coefficients \( C_k = C_{n-k} \).

**Hint 2:** Use the symmetry property of binomial coefficients to rewrite the terms.

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**Step 3: Rewrite \( S \).**
By reversing the indices, we can write:

\[ S = C_n^2 + 2C_{n-1}^2 + 3C_{n-2}^2 + \ldots + (n + 1)C_0^2 \]

**Hint 3:** This gives you a second expression for \( S \) which can be combined with the first.

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**Step 4: Add the two expressions for \( S \).**
Now, we have two expressions for \( S \):

\[ S = C_0^2 + 2C_1^2 + 3C_2^2 + \ldots + (n + 1)C_n^2 \]
\[ S = C_n^2 + 2C_{n-1}^2 + 3C_{n-2}^2 + \ldots + (n + 1)C_0^2 \]

Adding these two gives:

\[ 2S = (n + 1)(C_0^2 + C_1^2 + C_2^2 + \ldots + C_n^2) \]

**Hint 4:** Notice that you can factor out \( (n + 1) \) from the sum.

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**Step 5: Use the known formula for the sum of squares of binomial coefficients.**
The sum of the squares of the binomial coefficients is given by:

\[ C_0^2 + C_1^2 + C_2^2 + \ldots + C_n^2 = C_{2n}^0 \]

Thus, we can write:

\[ 2S = (n + 1) C_{2n}^0 \]

**Hint 5:** Remember the identity for the sum of squares of binomial coefficients.

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**Step 6: Solve for \( S \).**
Now, we can solve for \( S \):

\[ S = \frac{(n + 1)}{2} C_{2n}^0 \]

**Hint 6:** Simplify the expression to find the final result.

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**Final Result:**
The value of the original expression is:

\[ S = \frac{(n + 1)}{2} C_{2n}^0 \]

This gives us the required sum in terms of \( n \) and the binomial coefficients.

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