To solve this problem, we can treat José's savings as a geometric sequence. In a geometric sequence, each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.

Given:

• First day's savings ($a_1$): 5 pesos
• Common ratio ($r$): 2 (because he doubles his savings each day)

To Find:

• The amount saved on the 9th day ($a_9$)

Formula for the nth term of a geometric sequence:

$a_n = a_1 \cdot r^{(n-1)}$

Substituting the known values:

$a_9 = 5 \cdot 2^{(9-1)}$ $a_9 = 5 \cdot 2^{8}$

Calculate $2^8$:

$2^8 = 256$

Final calculation:

$a_9 = 5 \cdot 256 = 1280 \text{ pesos}$

Conclusion:

José will save 1280 pesos on the 9th day.

Would you like more details or have any questions?

Here are 5 related questions:

1. What will be the total savings after 9 days?
2. How would the sequence change if José saved 10 pesos on the first day instead?
3. What is the sum of José's savings from the 1st to the 9th day?
4. How many days will it take for José's savings to exceed 10,000 pesos?
5. If José stops saving after the 9th day, what would be the average amount saved per day?

Tip: In a geometric sequence, knowing the first term and the common ratio allows you to find any term in the sequence using the formula provided.