if `sum_(r=0)^(25).^(50)C_(r)(.^(50-r)C_(25-r))=k(.^(50)C_25)`, then k equals:

Options

A. `2^(25)`
B. `2^(25) -1`
C. `2^(24)`
D. `(25)^(2)`

Answer

To solve the problem, we need to evaluate the expression

\[ \sum_{r=0}^{25} \binom{50}{r} \binom{50-r}{25-r} \]

and express it in the form \( k \cdot \binom{50}{25} \).

### Step 1: Understand the Binomial Coefficient

Recall that the binomial coefficient \(\binom{n}{r}\) is defined as:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

### Step 2: Rewrite the Summation

We can rewrite the summation as follows:

\[ \sum_{r=0}^{25} \binom{50}{r} \binom{50-r}{25-r} \]

### Step 3: Simplify the Terms

Using the definition of the binomial coefficient, we have:

\[ \binom{50}{r} = \frac{50!}{r!(50-r)!} \]
and
\[ \binom{50-r}{25-r} = \frac{(50-r)!}{(25-r)!(50-25)!} = \frac{(50-r)!}{(25-r)! \cdot 25!} \]

### Step 4: Combine the Terms

Now, substituting these into the summation:

\[ \sum_{r=0}^{25} \frac{50!}{r!(50-r)!} \cdot \frac{(50-r)!}{(25-r)! \cdot 25!} \]

This simplifies to:

\[ \sum_{r=0}^{25} \frac{50!}{r!(25-r)! \cdot 25!} \]

### Step 5: Factor Out Constants

Notice that \( \frac{50!}{25!} \) is a common factor:

\[ \frac{50!}{25!} \sum_{r=0}^{25} \frac{1}{r!(25-r)!} \]

### Step 6: Recognize the Binomial Theorem

The summation \( \sum_{r=0}^{25} \frac{1}{r!(25-r)!} \) is the expansion of \( (1+1)^{25} \):

\[ \sum_{r=0}^{25} \binom{25}{r} = 2^{25} \]

### Step 7: Combine Everything Together

Thus, we can rewrite our expression as:

\[ \frac{50!}{25!} \cdot \frac{1}{25!} \cdot 2^{25} \]

This is equivalent to:

\[ \binom{50}{25} \cdot 2^{25} \]

### Step 8: Set Equal to Given Expression

Now, we have:

\[ \sum_{r=0}^{25} \binom{50}{r} \binom{50-r}{25-r} = \binom{50}{25} \cdot 2^{25} \]

According to the problem, this equals \( k \cdot \binom{50}{25} \).

### Step 9: Solve for \( k \)

By comparing both sides, we find:

\[ k = 2^{25} \]

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{2^{25}} \]