The value of `(1)/(1!50!)+(1)/(3!48!)+(1)/(5!46!)+......+(1)/(49!2!)+(1)/(51!1!)` is

Options

A. `2^(51)/(50!)`
B. `(2^(50))/(50!)`
C. `2^(51)/(51!)`
D. `(2^(50))/(51!)`

Answer

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

\[ S = \frac{1}{1!50!} + \frac{1}{3!48!} + \frac{1}{5!46!} + \ldots + \frac{1}{49!2!} + \frac{1}{51!1!} \]

### Step 1: Identify the Pattern
Notice that the denominators consist of factorials where the sum of the indices is always 51. This can be expressed as:

\[ S = \sum_{k=0}^{25} \frac{1}{(2k+1)!(51-2k)!} \]

### Step 2: Factor Out \(51!\)
We can factor out \(51!\) from the terms:

\[ S = \frac{1}{51!} \left( \sum_{k=0}^{25} \frac{51!}{(2k+1)!(51-2k)!} \right) \]

### Step 3: Recognize the Combinatorial Identity
The term inside the summation resembles the binomial coefficient:

\[ \frac{51!}{(2k+1)!(51-2k)!} = \binom{51}{2k+1} \]

Thus, we can rewrite the sum as:

\[ S = \frac{1}{51!} \sum_{k=0}^{25} \binom{51}{2k+1} \]

### Step 4: Use the Binomial Theorem
The sum of the binomial coefficients can be evaluated using the binomial theorem. The sum of all odd indexed binomial coefficients from \( \binom{n}{0} \) to \( \binom{n}{n} \) is given by:

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

For odd indices, it can be shown that:

\[ \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k+1} = \frac{1}{2} (2^n) = 2^{n-1} \]

For \( n = 51 \):

\[ \sum_{k=0}^{25} \binom{51}{2k+1} = 2^{50} \]

### Step 5: Substitute Back
Now substituting back into our expression for \( S \):

\[ S = \frac{1}{51!} \cdot 2^{50} \]

### Final Answer
Thus, the value of the original expression is:

\[ S = \frac{2^{50}}{51!} \]