The value of `((50),(0)) ((50),(1)) + ((50), (1)) ((50),(2)) +............+ ((50),(49))((50),(50)) `is

Options

A. `((100),(50))`
B. `((100),(51))`
C. `((50),(25))`
D. `((50),(25))^(2)`

Answer

To solve the problem, we need to evaluate the sum:

\[ \sum_{k=0}^{50} \binom{50}{k} \binom{50}{50-k} \]

This can be rewritten using the symmetry property of binomial coefficients:

\[ \sum_{k=0}^{50} \binom{50}{k} \binom{50}{k} \]

This is equivalent to:

\[ \sum_{k=0}^{50} \binom{50}{k}^2 \]

### Step 1: Use the Binomial Theorem

According to the binomial theorem, we know that:

\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

If we set \( n = 50 \) and \( x = 1 \), we get:

\[ (1 + 1)^{50} = 2^{50} = \sum_{k=0}^{50} \binom{50}{k} \]

### Step 2: Consider the Square of the Binomial Expansion

Now, consider the square of the binomial expansion:

\[ (1 + x)^{50} \cdot (1 + x)^{50} = (1 + x)^{100} \]

The left-hand side can be expressed as:

\[ \sum_{k=0}^{50} \binom{50}{k} x^k \cdot \sum_{j=0}^{50} \binom{50}{j} x^j \]

When we multiply these two expansions, the coefficient of \( x^{k} \) in the product will be:

\[ \sum_{j=0}^{k} \binom{50}{j} \binom{50}{k-j} \]

This is equal to \( \binom{100}{k} \) (by Vandermonde's identity).

### Step 3: Find the Coefficient of \( x^{50} \)

To find the sum \( \sum_{k=0}^{50} \binom{50}{k}^2 \), we need to find the coefficient of \( x^{50} \) in \( (1 + x)^{100} \):

\[ (1 + x)^{100} = \sum_{k=0}^{100} \binom{100}{k} x^k \]

The coefficient of \( x^{50} \) is:

\[ \binom{100}{50} \]

### Step 4: Conclusion

Thus, we conclude that:

\[ \sum_{k=0}^{50} \binom{50}{k} \binom{50}{k} = \binom{100}{50} \]

### Final Answer

The value of the given expression is:

\[ \binom{100}{50} = 100891344545564193334812497256 \]