The value of sum `(""^(n) C_(1) )^(2)+ (""^(n) C_(2) )^(2) + (""^(n) C_(3))^(2) + …+ (""^(n) C_(n) )^(2)` is

Options

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

Answer

To find the value of the sum \( \sum_{k=1}^{n} \binom{n}{k}^2 \), we can use the properties of binomial coefficients and the Binomial Theorem. Here’s a step-by-step solution:

### Step 1: Understand the Binomial Coefficient
The binomial coefficient \( \binom{n}{k} \) represents the number of ways to choose \( k \) elements from a set of \( n \) elements. The square of the binomial coefficient \( \binom{n}{k}^2 \) can be interpreted combinatorially.

### Step 2: Use the Binomial Theorem
The Binomial Theorem states that:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
We can apply this theorem to find the sum of squares of the binomial coefficients.

### Step 3: Expand \( (1 + x)^n \) and \( (1 + x)^n \)
Consider the expression:
\[ (1 + x)^n \cdot (1 + x)^n = (1 + x)^{2n} \]
This expands to:
\[ \sum_{k=0}^{2n} \binom{2n}{k} x^k \]
On the left side, we can express it as:
\[ \left( \sum_{k=0}^{n} \binom{n}{k} x^k \right)^2 \]
### Step 4: Compare Coefficients
When we expand \( \left( \sum_{k=0}^{n} \binom{n}{k} x^k \right)^2 \), the coefficient of \( x^n \) will be:
\[ \sum_{k=0}^{n} \binom{n}{k}^2 \]
This means:
\[ \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \]

### Step 5: Calculate the Required Sum
Since we want the sum from \( k=1 \) to \( n \):
\[ \sum_{k=1}^{n} \binom{n}{k}^2 = \binom{2n}{n} - \binom{n}{0}^2 \]
Given that \( \binom{n}{0} = 1 \):
\[ \sum_{k=1}^{n} \binom{n}{k}^2 = \binom{2n}{n} - 1 \]

### Final Result
Thus, the value of the sum \( \sum_{k=1}^{n} \binom{n}{k}^2 \) is:
\[ \boxed{\binom{2n}{n} - 1} \]