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 $|r| < 1$.