Let's break down Perry's carrot consumption day by day.
- Day 1: Perry starts with 20 carrots.
- Day 2: He eats half of the 20 carrots, so he eats 10 and has \(20 - 10 = 10\) left.
- Day 3: He eats half of the 10 carrots, so he eats 5 and has \(10 - 5 = 5\) left.
- Day 4: He eats half of the 5 carrots, so he eats 2.5 (in practical terms, this means he might end up with 2 or 3, but we can use decimals for the formula) and has \(5 - 2.5 = 2.5\) left.
- Day 5: He eats half of 2.5, which is 1.25, and has \(2.5 - 1.25 = 1.25\) left.
- Continuing this process, we see that each day he eats half of what is remaining.
To express this mathematically, we can define a sequence where the number of carrots left on day \(n\) can be represented with the following formula:
- Let \(C(n)\) be the number of carrots left on day \(n\).
- The initial count on day 1 (when \( n = 1 \)) is \(C(1) = 20\).
- Each subsequent day, the number of carrots left is half of what was left the previous day: \[ C(n) = \frac{C(n-1)}{2} \]
Using this recursive definition, we can express the number of carrots on any given day \(n\): \[ C(n) = 20 \times \left(\frac{1}{2}\right)^{n-1} \]
This formula shows that the number of carrots left on day \(n\) is equal to the initial number of carrots (20) multiplied by \(\frac{1}{2}\) raised to the power of \(n-1\).
Thus, the complete sequence formula describing the number of carrots Perry will have on any given day \(n\) is: \[ C(n) = 20 \times \left(\frac{1}{2}\right)^{n-1} \]