If `A=4^(12),B=1+4+4^(2)+4^(3)+...+4^(11),C=1+16+16^(2)+...+16^(5)` then which of the following is True ?

Options

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

Answer

To solve the problem, we need to evaluate the expressions for A, B, and C step by step.

### Step 1: Calculate A
Given:
\[ A = 4^{12} \]

This is already in its simplest form.

### Step 2: Calculate B
The expression for B is:
\[ B = 1 + 4 + 4^2 + 4^3 + ... + 4^{11} \]

This is a geometric series where:
- First term \( a = 1 \)
- Common ratio \( r = 4 \)
- Number of terms \( n = 12 \) (from \( 4^0 \) to \( 4^{11} \))

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

Substituting the values:
\[ B = \frac{1(4^{12} - 1)}{4 - 1} = \frac{4^{12} - 1}{3} \]

### Step 3: Calculate C
The expression for C is:
\[ C = 1 + 16 + 16^2 + 16^3 + 16^4 + 16^5 \]

This is also a geometric series where:
- First term \( a = 1 \)
- Common ratio \( r = 16 \)
- Number of terms \( n = 6 \) (from \( 16^0 \) to \( 16^5 \))

Using the same formula for the sum of a geometric series:
\[ C = \frac{1(16^6 - 1)}{16 - 1} = \frac{16^6 - 1}{15} \]

### Step 4: Compare A, B, and C
Now we have:
- \( A = 4^{12} \)
- \( B = \frac{4^{12} - 1}{3} \)
- \( C = \frac{16^6 - 1}{15} \)

We can express \( 16^6 \) in terms of \( 4^{12} \):
\[ 16^6 = (4^2)^6 = 4^{12} \]
Thus:
\[ C = \frac{4^{12} - 1}{15} \]

### Step 5: Determine the relationship between A, B, and C
Now we compare the values:
1. \( A = 4^{12} \)
2. \( B = \frac{4^{12} - 1}{3} \)
3. \( C = \frac{4^{12} - 1}{15} \)

Since \( 4^{12} \) is a large number, we can see that:
- \( B \) is slightly less than \( A \)
- \( C \) is significantly less than both \( A \) and \( B \)

Thus, the order from largest to smallest is:
\[ A > B > C \]

### Conclusion
The correct relationship is:
- A is the largest,
- B is the next,
- C is the smallest.