Find the sum of the series `sum_(r=0)^(50)(""^(100-r)C_(25))/(100-r)`

Options

A. `(1)/(25)[""^(100)C_(25)+""^(50)C_(25)]`
B. `(1)/(25)[""^(100)C_(25)-""^(51)C_(25)]`
C. `(1)/(25)[""^(100)C_(25)-""^(49)C_(25)]`
D. `(1)/(25)[""^(100)C_(25)-""^(50)C_(26)]`

Explanation

To find the sum of the series

\[ \sum_{r=0}^{50} \frac{\binom{100-r}{25}}{100-r}, \]

we can follow these steps:

### Step 1: Rewrite the binomial coefficient
We can use the property of binomial coefficients:

\[ \binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}. \]

Applying this to our series, we have:

\[ \binom{100-r}{25} = \frac{100-r}{25} \binom{99-r}{24}. \]

### Step 2: Substitute into the series
Substituting this into the series gives:

\[ \sum_{r=0}^{50} \frac{\frac{100-r}{25} \binom{99-r}{24}}{100-r} = \sum_{r=0}^{50} \frac{1}{25} \binom{99-r}{24}. \]

### Step 3: Factor out the constant
Since \(\frac{1}{25}\) is a constant, we can factor it out of the summation:

\[ \frac{1}{25} \sum_{r=0}^{50} \binom{99-r}{24}. \]

### Step 4: Change the index of summation
To simplify the summation, we can change the index of summation. Let \(k = 50 - r\), then when \(r = 0\), \(k = 50\) and when \(r = 50\), \(k = 0\). Thus, the summation becomes:

\[ \sum_{k=0}^{50} \binom{99 - (50 - k)}{24} = \sum_{k=0}^{50} \binom{49 + k}{24}. \]

### Step 5: Use the Hockey Stick Identity
Using the Hockey Stick Identity (or Christmas Stocking Theorem), we have:

\[ \sum_{j=a}^{b} \binom{j}{r} = \binom{b+1}{r+1}. \]

In our case, we can apply it to get:

\[ \sum_{k=0}^{50} \binom{49 + k}{24} = \binom{50 + 49 + 1}{24 + 1} = \binom{100}{25}. \]

### Step 6: Combine results
Now substituting back into our expression gives:

\[ \frac{1}{25} \cdot \binom{100}{25}. \]

### Final Result
Thus, the sum of the series is:

\[ \frac{1}{25} \cdot \binom{100}{25}. \]