The numbers of bacteria in certain doubles every hour. If there were 25 bacteria present originally, how many bacteria will be present at the end of `6^(th)` hour ?

Options

A. 800
B. 1600
C. `2(25)^(6)`
D. `2(25)^(5)`

Answer

To solve the problem of how many bacteria will be present at the end of the 6th hour when the initial count is 25 and the number doubles every hour, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Number of Bacteria**:
The problem states that the initial number of bacteria is 25.
\[ A_1 = 25 \]

2. **Determine the Doubling Factor**:
Since the bacteria double every hour, the common ratio \( r \) is 2.
\[ r = 2 \]

3. **Find the Formula for the Number of Bacteria**:
The number of bacteria after \( n \) hours can be expressed using the formula for the \( n \)-th term of a geometric progression:
\[ A_n = A_1 \cdot r^{(n-1)} \]
Here, we need to find the number of bacteria at the beginning of the 7th hour, which corresponds to \( n = 7 \).

4. **Substituting Values into the Formula**:
We substitute \( A_1 = 25 \), \( r = 2 \), and \( n = 7 \) into the formula:
\[ A_7 = 25 \cdot 2^{(7-1)} = 25 \cdot 2^6 \]

5. **Calculate \( 2^6 \)**:
Calculate \( 2^6 \):
\[ 2^6 = 64 \]

6. **Final Calculation**:
Now, multiply the initial number of bacteria by \( 2^6 \):
\[ A_7 = 25 \cdot 64 = 1600 \]

7. **Conclusion**:
Therefore, the number of bacteria present at the beginning of the 7th hour is:
\[ \text{Number of bacteria} = 1600 \]