The value of ` 99^(50) - 99.98^(50) + (99*98)/(1*2) (97)^(50) -…+ 99 ` is

Answer

To solve the expression \( 99^{50} - 99 \cdot 98^{50} + \frac{99 \cdot 98}{1 \cdot 2} \cdot 97^{50} - \ldots + 99 \), we can utilize the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the Pattern**:
The expression can be rewritten as:
\[ S = 99^{50} - 99 \cdot 98^{50} + \frac{99 \cdot 98}{1 \cdot 2} \cdot 97^{50} - \ldots + 99 \]
This resembles the expansion of \( (99 - 1)^{50} \) using the Binomial Theorem.

2. **Apply the Binomial Theorem**:
According to the Binomial Theorem, we have:
\[ (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \]
Here, let \( a = 99 \), \( b = 1 \), and \( n = 50 \):
\[ (99 - 1)^{50} = \sum_{k=0}^{50} \binom{50}{k} 99^{50-k} (-1)^k \]

3. **Rearranging the Terms**:
The expression \( (99 - 1)^{50} \) simplifies to:
\[ 98^{50} = \sum_{k=0}^{50} \binom{50}{k} 99^{50-k} (-1)^k \]
This means that the terms in our original expression correspond to the coefficients in this expansion.

4. **Evaluate the Expression**:
The original expression \( S \) can be interpreted as the sum of the first 51 terms of the expansion of \( (99 - 1)^{50} \). Thus:
\[ S = (99 - 1)^{50} = 98^{50} \]

5. **Final Calculation**:
Since \( 98^{50} \) is a positive number, and the alternating sum of the binomial coefficients leads to cancellation of terms, we find that:
\[ S = 0 \]

### Conclusion:
The value of the expression \( 99^{50} - 99 \cdot 98^{50} + \frac{99 \cdot 98}{1 \cdot 2} \cdot 97^{50} - \ldots + 99 \) is:
\[ \boxed{0} \]