`sqrt(5){(sqrt(5)+1)^(50)-(sqrt(5)-1)^(50)}`

Options

A. An irrational number
B. 0
C. A natural number
D. A prime number

Answer

To solve the expression \(\sqrt{5} \left( (\sqrt{5}+1)^{50} - (\sqrt{5}-1)^{50} \right)\), we will use the Binomial Theorem.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We need to evaluate \(\sqrt{5} \left( (\sqrt{5}+1)^{50} - (\sqrt{5}-1)^{50} \right)\).

2. **Applying Binomial Theorem**:
According to the Binomial Theorem, we can expand \((a+b)^n\) as:
\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
Here, we will expand both \((\sqrt{5}+1)^{50}\) and \((\sqrt{5}-1)^{50}\).

3. **Expanding \((\sqrt{5}+1)^{50}\)**:
\[ (\sqrt{5}+1)^{50} = \sum_{k=0}^{50} \binom{50}{k} (\sqrt{5})^{50-k} (1)^k = \sum_{k=0}^{50} \binom{50}{k} (\sqrt{5})^{50-k} \]

4. **Expanding \((\sqrt{5}-1)^{50}\)**:
\[ (\sqrt{5}-1)^{50} = \sum_{k=0}^{50} \binom{50}{k} (\sqrt{5})^{50-k} (-1)^k \]

5. **Combining the Expansions**:
Now, we subtract the two expansions:
\[ (\sqrt{5}+1)^{50} - (\sqrt{5}-1)^{50} = \sum_{k=0}^{50} \binom{50}{k} (\sqrt{5})^{50-k} - \sum_{k=0}^{50} \binom{50}{k} (\sqrt{5})^{50-k} (-1)^k \]
This simplifies to:
\[ = \sum_{k=0}^{50} \binom{50}{k} (\sqrt{5})^{50-k} (1 - (-1)^k) \]

6. **Identifying Odd and Even Terms**:
The term \(1 - (-1)^k\) is zero for even \(k\) and \(2\) for odd \(k\). Thus, we only consider the odd \(k\):
\[ = 2 \sum_{k \text{ odd}} \binom{50}{k} (\sqrt{5})^{50-k} \]

7. **Final Expression**:
Therefore, we have:
\[ \sqrt{5} \left( (\sqrt{5}+1)^{50} - (\sqrt{5}-1)^{50} \right) = 2\sqrt{5} \sum_{k \text{ odd}} \binom{50}{k} (\sqrt{5})^{50-k} \]

8. **Simplifying the Sum**:
The sum can be expressed as:
\[ = 2\sqrt{5} \left( \sum_{j=0}^{25} \binom{50}{2j+1} (\sqrt{5})^{50-(2j+1)} \right) \]
where \(j\) runs from \(0\) to \(25\).

### Conclusion:
The final result is:
\[ \sqrt{5} \left( (\sqrt{5}+1)^{50} - (\sqrt{5}-1)^{50} \right) = 2\sqrt{5} \sum_{k \text{ odd}} \binom{50}{k} (\sqrt{5})^{50-k} \]