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To solve this problem, we will compute the following sums step by step, using appropriate formulas for summation.

### (i) \( \sum_{k=1}^{6} 4^k \)

This is a geometric series where the first term \( a = 4^1 = 4 \) and the common ratio \( r = 4 \). The sum of the first \( n \) terms of a geometric series is given by:

\[
S_n = \frac{a(r^n - 1)}{r - 1}
\]

Here, \( n = 6 \), so:

\[
S_6 = \frac{4(4^6 - 1)}{4 - 1} = \frac{4(4096 - 1)}{3} = \frac{4(4095)}{3} = \frac{16380}{3} = 5460
\]

Thus, the sum for part (i) is **5460**.

---

### (ii) \( \sum_{k=1}^{6} (0.25)^k \) (correct to 6 decimal places)

This is another geometric series where \( a = 0.25^1 = 0.25 \) and \( r = 0.25 \). Using the geometric sum formula again:

\[
S_n = \frac{a(1 - r^n)}{1 - r}
\]

For \( n = 6 \), \( a = 0.25 \), and \( r = 0.25 \):

\[
S_6 = \frac{0.25(1 - (0.25)^6)}{1 - 0.25}
\]

First, calculate \( (0.25)^6 \):

\[
(0.25)^6 = 0.000244140625
\]

Now plug the values into the formula:

\[
S_6 = \frac{0.25(1 - 0.000244140625)}{0.75} = \frac{0.25(0.999755859375)}{0.75} = \frac{0.24993896484375}{0.75} \approx 0.333251953125
\]

Rounding to 6 decimal places, the sum for part (ii) is **0.333252**.

---

### (iii) \( \sum_{k=1}^{\infty} (0.25)^k \)

This is an infinite geometric series where \( a = 0.25 \) and \( r = 0.25 \). The sum of an infinite geometric series is given by:

\[
S_\infty = \frac{a}{1 - r}
\]

Here, \( a = 0.25 \) and \( r = 0.25 \):

\[
S_\infty = \frac{0.25}{1 - 0.25} = \frac{0.25}{0.75} = \frac{1}{3} \approx 0.333333
\]

Thus, the sum for part (iii) is **0.333333**.

---

Let me know if you would like further explanations on any part!

### Related questions:
1. What is the general formula for the sum of a geometric series?
2. How does the sum of an infinite geometric series differ from a finite geometric series?
3. What happens when the common ratio in a geometric series is greater than 1?
4. Can you calculate the sum of an arithmetic series using a similar formula?
5. How do geometric series appear in real-world applications like economics or physics?

**Tip**: The sum of an infinite geometric series only converges when the absolute value of the common ratio \( |r| < 1 \).

Calculate the following sums using the appropriate formulas: (i) Sum of 4^k from k=1 to 6, (ii) Sum of (0.25)^k from k=1 to 6 (correct to 6 decimal places), (iii) Sum of (0.25)^k from k=1 to infinity.

Explore how to calculate the sums of geometric series like 4^k and (0.25)^k using formulas. This problem covers key topics such as finite and infinite geometric series and is suitable for students in Grades 9-12.

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