Let `P = sum_(r=1)^(50) (""^(50+r)C_(r)(2r-1))/(""^(50)C_(r)(50+r)), Q = sum__(r=1)^(50) (""^(50)C_(r))^(2), R = sum_(r=0)^(100) (-1)^(r) (""^(100)C_(r))^(2)`
The value of `P - Q` is equal to

Explanation

To solve the problem, we need to calculate the values of \( P \), \( Q \), and then find \( P - Q \).

### Step-by-Step Solution

1. **Calculate \( P \)**:
\[ P = \sum_{r=1}^{50} \frac{{\binom{50+r}{r} (2r-1)}}{{\binom{50}{r} (50+r)}} \]
We can rewrite \( P \) using the identity \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). This gives us:
\[ P = \sum_{r=1}^{50} \frac{{(50+r)!}}{{r!(50)!}} \cdot (2r-1) \cdot \frac{(50)!}{(50+r)!} \]
Simplifying this, we get:
\[ P = \sum_{r=1}^{50} \frac{(2r-1)}{(50+r)} \cdot \binom{50+r}{r} \]

2. **Rearranging \( P \)**:
We can express \( P \) as:
\[ P = \sum_{r=1}^{50} \left( \binom{50+r}{r} - \binom{49+r}{r-1} \right) \]
This can be simplified further to:
\[ P = \binom{100}{50} - 1 \]

3. **Calculate \( Q \)**:
\[ Q = \sum_{r=1}^{50} \binom{50}{r}^2 \]
Using the combinatorial identity:
\[ \sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n} \]
We have:
\[ Q = \binom{100}{50} \]

4. **Calculate \( P - Q \)**:
Now we can find \( P - Q \):
\[ P - Q = \left( \binom{100}{50} - 1 \right) - \binom{100}{50} \]
Simplifying this gives:
\[ P - Q = -1 \]

### Final Answer
Thus, the value of \( P - Q \) is:
\[ \boxed{-1} \]