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

Answer

To solve the problem \( 99^{50} - 99 \cdot 98^{50} + \frac{99 \cdot 98}{2} \cdot 97^{50} - \ldots + 99 \), we will use the Binomial Theorem and properties of binomial coefficients.

### Step-by-Step Solution:

1. **Identify the Pattern**:
The expression can be rewritten in terms of binomial coefficients. Notice that the expression alternates in signs and involves powers of decreasing integers. This suggests a binomial expansion.

2. **Rewrite the Expression**:
The expression can be expressed as:
\[ S = \sum_{k=0}^{99} (-1)^k \binom{99}{k} (99-k)^{50} \]
This means we are summing over all \( k \) from 0 to 99, where \( \binom{99}{k} \) is the binomial coefficient.

3. **Apply the Binomial Theorem**:
According to the Binomial Theorem, we have:
\[ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
In our case, we can set \( x = 99 \) and \( y = -1 \) and \( n = 50 \):
\[ (99 - 1)^{50} = 98^{50} \]
This gives us:
\[ 98^{50} = \sum_{k=0}^{50} \binom{50}{k} 99^{50-k} (-1)^k \]

4. **Relate the Two Expressions**:
We can relate our sum \( S \) to the binomial expansion:
\[ S = 99^{50} - 98^{50} \]
This is because the sum \( S \) effectively computes the value of \( (99-1)^{50} \) using the alternating signs.

5. **Simplify the Expression**:
Now, we can simplify \( S \):
\[ S = 99^{50} - 98^{50} \]

6. **Final Calculation**:
The value \( 99^{50} - 98^{50} \) can be computed directly or left in this form, as it represents the difference of two powers.

### Final Answer:
The value of \( 99^{50} - 99 \cdot 98^{50} + \frac{99 \cdot 98}{2} \cdot 97^{50} - \ldots + 99 \) is:
\[ 99^{50} - 98^{50} \]