Larger of `99^(50) +100^(50) and 101^(50)` is ………..

Answer

To determine which is larger between \( 99^{50} + 100^{50} \) and \( 101^{50} \), we can use the Binomial Theorem to expand \( 101^{50} \) and compare it with the sum \( 99^{50} + 100^{50} \).

### Step-by-Step Solution:

1. **Express \( 101^{50} \) using the Binomial Theorem**:
\[ 101^{50} = (100 + 1)^{50} \]
By applying the Binomial Theorem, we can expand this:
\[ 101^{50} = \sum_{k=0}^{50} \binom{50}{k} 100^{50-k} \cdot 1^k \]
This gives us:
\[ 101^{50} = \binom{50}{0} 100^{50} + \binom{50}{1} 100^{49} + \binom{50}{2} 100^{48} + \ldots + \binom{50}{50} 1 \]

2. **Identify the first few terms of the expansion**:
The first few terms of the expansion are:
\[ 100^{50} + 50 \cdot 100^{49} + \frac{50 \cdot 49}{2} \cdot 100^{48} + \ldots + 1 \]
Thus,
\[ 101^{50} = 100^{50} + 50 \cdot 100^{49} + \text{(other positive terms)} \]

3. **Express \( 99^{50} \) using the Binomial Theorem**:
Similarly, we can express \( 99^{50} \) as:
\[ 99^{50} = (100 - 1)^{50} \]
Expanding this using the Binomial Theorem gives:
\[ 99^{50} = \sum_{k=0}^{50} \binom{50}{k} 100^{50-k} (-1)^k \]
The first few terms are:
\[ 100^{50} - 50 \cdot 100^{49} + \frac{50 \cdot 49}{2} \cdot 100^{48} - \ldots + (-1)^{50} \]

4. **Combine \( 99^{50} + 100^{50} \)**:
Now, we can combine \( 99^{50} + 100^{50} \):
\[ 99^{50} + 100^{50} = \left(100^{50} - 50 \cdot 100^{49} + \frac{50 \cdot 49}{2} \cdot 100^{48} - \ldots \right) + 100^{50} \]
This simplifies to:
\[ 2 \cdot 100^{50} - 50 \cdot 100^{49} + \frac{50 \cdot 49}{2} \cdot 100^{48} - \ldots \]

5. **Compare \( 99^{50} + 100^{50} \) and \( 101^{50} \)**:
Now we compare:
\[ 99^{50} + 100^{50} = 2 \cdot 100^{50} - 50 \cdot 100^{49} + \ldots \]
with
\[ 101^{50} = 100^{50} + 50 \cdot 100^{49} + \ldots \]
The key observation is that the negative terms in \( 99^{50} + 100^{50} \) will reduce the total, while \( 101^{50} \) has positive contributions from its expansion.

6. **Conclusion**:
Since \( 101^{50} \) has more positive contributions and fewer negative terms compared to \( 99^{50} + 100^{50} \), we conclude:
\[ 101^{50} > 99^{50} + 100^{50} \]

Thus, the larger of \( 99^{50} + 100^{50} \) and \( 101^{50} \) is:
\[ \boxed{101^{50}} \]