
Performance Task
13
Donation for a Cause
1. You are running a foundation for street children. Mr. Felipe Cortejos, a rich benefactor, offers two options for his donations: Option A: To give Php 1 000 on day 1, Php 999 on day 2, Php 998 on day 3, with the process to end
after 1 000 days.
Option B: To give Php 1 on day 1, Php 2 on day 2, Php 4 on day 3, for 1 000 days.
You have to tell him today which option you want to take. Which of the two options results in more money for the foundation? Show your computations.
 Use An = A1 + (n-1)d on both options

To determine which donation option results in more money for the foundation, we need to compute the total amount of money received for each option.

### **Option A: Arithmetic Sequence**

- **First term (A₁) = Php 1,000**
- **Common difference (d) = -1**
- **Number of terms (n) = 1,000**

In an arithmetic sequence, the sum of the first \( n \) terms is given by the formula:

\[
S_n = \frac{n}{2} \times (2A_1 + (n-1)d)
\]

Substituting the values:

\[
S_{1000} = \frac{1000}{2} \times (2 \times 1000 + (1000-1) \times (-1))
\]

\[
S_{1000} = 500 \times (2000 + 999 \times (-1))
\]

\[
S_{1000} = 500 \times (2000 - 999) = 500 \times 1001 = 500,500
\]

### **Option B: Geometric Sequence**

- **First term (A₁) = Php 1**
- **Common ratio (r) = 2**
- **Number of terms (n) = 1,000**

In a geometric sequence, the sum of the first \( n \) terms is given by the formula:

\[
S_n = A_1 \times \frac{r^n - 1}{r - 1}
\]

Substituting the values:

\[
S_{1000} = 1 \times \frac{2^{1000} - 1}{2 - 1} = 2^{1000} - 1
\]

Since \( 2^{1000} \) is an extremely large number, let’s calculate its approximate value.

### Calculating \( 2^{1000} \)

We know that:

\[
2^{10} = 1024 \approx 10^3
\]

So,

\[
2^{1000} = (2^{10})^{100} \approx 10^{300}
\]

Therefore, \( 2^{1000} - 1 \) is approximately \( 10^{300} \), which is an astronomically large number compared to the sum for Option A.

### **Conclusion**
- **Option A:** The total donation is Php 500,500.
- **Option B:** The total donation is approximately \( 10^{300} \) pesos, which is far greater than Php 500,500.

Thus, **Option B** results in significantly more money for the foundation.

Would you like more details, or do you have any further questions?

### Related Questions:
1. How does the sum formula for geometric series change if the common ratio is less than 1?
2. What happens if the donation period is extended to 2,000 days? Which option would be better then?
3. Can we find the exact value of \( 2^{1000} - 1 \)? How do super-large numbers affect real-world calculations?
4. What if the common difference in Option A was different, like -2 or -3? How would that change the outcome?
5. How can this concept of arithmetic and geometric sequences be applied to financial planning or investment strategies?

### Tip:
For problems involving extremely large numbers, it's often easier to approximate their scale rather than calculate them exactly, as the differences can be enormous.

You are running a foundation for street children. Mr. Felipe Cortejos, a rich benefactor, offers two options for his donations: Option A: To give Php 1 000 on day 1, Php 999 on day 2, Php 998 on day 3, with the process to end after 1 000 days. Option B: To give Php 1 on day 1, Php 2 on day 2, Php 4 on day 3, for 1 000 days. You have to tell him today which option you want to take. Which of the two options results in more money for the foundation? Show your computations.

Explore which donation option yields more money for a foundation for street children. This problem involves arithmetic and geometric sequences, calculating sums over 1,000 days. Suitable for high school students in Grades 10-12, this task covers essential algebraic concepts and formulas.

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