By using binomial theorem find which number is greater `(1.2)^(3000) " or " 600`?

Explanation

To determine which number is greater, \( (1.2)^{3000} \) or \( 600 \), we can use the Binomial Theorem. Here’s a step-by-step solution:

### Step 1: Rewrite \( (1.2)^{3000} \)
We can express \( 1.2 \) as \( 1 + 0.2 \). Therefore, we rewrite the expression:
\[ (1.2)^{3000} = (1 + 0.2)^{3000} \]

### Step 2: Apply the Binomial Theorem
According to the Binomial Theorem, we can expand \( (a + b)^n \) as follows:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \( a = 1 \), \( b = 0.2 \), and \( n = 3000 \). Thus, we have:
\[ (1 + 0.2)^{3000} = \sum_{k=0}^{3000} \binom{3000}{k} (1)^{3000-k} (0.2)^k \]
This simplifies to:
\[ (1 + 0.2)^{3000} = \binom{3000}{0} (1)^{3000} (0.2)^0 + \binom{3000}{1} (1)^{2999} (0.2)^1 + \binom{3000}{2} (1)^{2998} (0.2)^2 + \ldots \]

### Step 3: Calculate the first few terms
Calculating the first few terms:
1. For \( k = 0 \):
\[ \binom{3000}{0} (1)^{3000} (0.2)^0 = 1 \]
2. For \( k = 1 \):
\[ \binom{3000}{1} (1)^{2999} (0.2)^1 = 3000 \times 0.2 = 600 \]
3. For \( k = 2 \):
\[ \binom{3000}{2} (1)^{2998} (0.2)^2 = \frac{3000 \times 2999}{2} \times 0.04 \]
This term is positive and contributes to the sum.

### Step 4: Sum of the terms
The sum of the first two terms is:
\[ 1 + 600 = 601 \]
Since the subsequent terms (for \( k \geq 2 \)) are all positive, we can conclude:
\[ (1.2)^{3000} = 1 + 600 + \text{(positive terms)} > 601 \]

### Step 5: Compare with 600
Since \( 601 > 600 \) and the additional positive terms only increase the total, we have:
\[ (1.2)^{3000} > 600 \]

### Conclusion
Thus, we conclude that:
\[ (1.2)^{3000} \text{ is greater than } 600 \]