If ` A = 2^(32) , B = 2^(31) + 2^(30) + 2 ^(29) + . . . + 2^(@) and C = 3 ^(15) + 3^(14) + 3^(13) + . . . + 3^(@) `, then which of the following options is TRUE

Options

A. `C gt B gt A `
B. ` C gt A gt A `
C. `A gt B gt C `
D. ` A gt C gt B `

Answer

To solve the problem, we need to evaluate the values of A, B, and C, and then compare them.

### Step 1: Evaluate A
Given:
\[ A = 2^{32} \]

### Step 2: Evaluate B
B is defined as:
\[ B = 2^{31} + 2^{30} + 2^{29} + \ldots + 2^{0} \]

This is a geometric series where:
- The first term \( a = 2^{31} \)
- The common ratio \( r = \frac{1}{2} \)
- The number of terms \( n = 32 \) (from \( 2^{31} \) to \( 2^{0} \))

The sum of a geometric series can be calculated using the formula:
\[ S_n = a \frac{1 - r^n}{1 - r} \]

Substituting the values:
\[ B = 2^{31} \cdot \frac{1 - \left(\frac{1}{2}\right)^{32}}{1 - \frac{1}{2}} = 2^{31} \cdot \frac{1 - \frac{1}{2^{32}}}{\frac{1}{2}} = 2^{31} \cdot 2 \left(1 - \frac{1}{2^{32}}\right) \]
\[ B = 2^{32} \left(1 - \frac{1}{2^{32}}\right) = 2^{32} - 1 \]

### Step 3: Evaluate C
C is defined as:
\[ C = 3^{15} + 3^{14} + 3^{13} + \ldots + 3^{0} \]

This is also a geometric series where:
- The first term \( a = 3^{15} \)
- The common ratio \( r = \frac{1}{3} \)
- The number of terms \( n = 16 \) (from \( 3^{15} \) to \( 3^{0} \))

Using the geometric series formula:
\[ C = 3^{15} \cdot \frac{1 - \left(\frac{1}{3}\right)^{16}}{1 - \frac{1}{3}} = 3^{15} \cdot \frac{1 - \frac{1}{3^{16}}}{\frac{2}{3}} = 3^{15} \cdot \frac{3}{2} \left(1 - \frac{1}{3^{16}}\right) \]
\[ C = \frac{3^{16}}{2} \left(1 - \frac{1}{3^{16}}\right) = \frac{3^{16} - 1}{2} \]

### Step 4: Compare A, B, and C
Now we have:
- \( A = 2^{32} \)
- \( B = 2^{32} - 1 \)
- \( C = \frac{3^{16} - 1}{2} \)

Now we will compare:
1. **Comparing A and B:**
\[ A = 2^{32} > B = 2^{32} - 1 \]

2. **Comparing B and C:**
We need to evaluate \( 2^{32} - 1 \) and \( \frac{3^{16} - 1}{2} \).

To compare these, we can multiply both sides by 2 (since both are positive):
\[ 2(2^{32} - 1) > 3^{16} - 1 \]
\[ 2^{33} - 2 > 3^{16} - 1 \]
\[ 2^{33} - 1 > 3^{16} \]

We know \( 2^{10} \approx 1024 \) and \( 3^5 = 243 \). Thus, \( 3^{16} = (3^5)^3 \cdot 3 \approx 243^3 \cdot 3 \), which is much smaller than \( 2^{33} \).

Therefore, we conclude that:
\[ B > C \]

### Conclusion
Putting it all together:
\[ A > B > C \]

### Final Answer
Thus, the correct option is that A is greater than B, and B is greater than C.

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