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

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

we can follow these steps:

### Step 1: Rewrite the Binomial Coefficient
We start by rewriting the binomial coefficient \({{100-r} \choose 25}\) using the identity:

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

Applying this to our series, we have:

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

### Step 2: Substitute into the Series
Substituting this back into the series gives us:

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

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

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

### 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, we can rewrite the sum:

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

### Step 5: Recognize the New Summation
The new summation \(\sum_{k=0}^{50} {49+k \choose 24}\) can be interpreted using the hockey-stick identity in combinatorics, which states:

\[ \sum_{j=r}^{n} {j \choose r} = {n+1 \choose r+1}. \]

In our case, we have:

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

### Step 6: Substitute Back into the Expression
Now substituting this back into our earlier expression gives:

\[ \frac{1}{25} {100 \choose 25}. \]

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

\[ \frac{1}{25} {100 \choose 25}. \]