To find the explicit formulas for the given recurrences using iteration, let's solve them step by step.

i. $T(n) = 3T\left(\frac{n}{3}\right) + n$, with $n = 3^k$ and $T(1) = 1$

Let's iterate the recurrence:

1. First iteration: $T(n) = 3T\left(\frac{n}{3}\right) + n$

2. Second iteration: $T\left(\frac{n}{3}\right) = 3T\left(\frac{n}{9}\right) + \frac{n}{3}$ Substituting into the first equation: $T(n) = 3\left(3T\left(\frac{n}{9}\right) + \frac{n}{3}\right) + n = 9T\left(\frac{n}{9}\right) + 3n + n = 9T\left(\frac{n}{9}\right) + 4n$

3. Third iteration: $T\left(\frac{n}{9}\right) = 3T\left(\frac{n}{27}\right) + \frac{n}{9}$ Substituting into the previous expression: $T(n) = 9\left(3T\left(\frac{n}{27}\right) + \frac{n}{9}\right) + 4n = 27T\left(\frac{n}{27}\right) + 9n + 4n = 27T\left(\frac{n}{27}\right) + 13n$

Continuing this pattern, we observe that after $m$ iterations:

$T(n) = 3^m T\left(\frac{n}{3^m}\right) + n \left(\sum_{i=0}^{m-1} 3^i\right)$

Since $n = 3^k$ and $T(1) = 1$, we have $\frac{n}{3^k} = 1$ when $m = k$:

$T(n) = 3^k T(1) + n \left(\sum_{i=0}^{k-1} 3^i\right)$

$T(n) = 3^k \cdot 1 + n \left(\frac{3^k - 1}{2}\right)$

$T(n) = 3^k + \frac{n(3^k - 1)}{2}$

Since $n = 3^k$, $3^k = n$:

$T(n) = n + \frac{n(n - 1)}{2} = \frac{n^2 + n}{2}$

Thus, the explicit formula is:

$\boxed{T(n) = \frac{n^2 + n}{2}}$

ii. $T(n) = 4T\left(\frac{n}{4}\right) + n$, with $n = 4^k$ and $T(1) = 1$

Let's iterate the recurrence:

1. First iteration: $T(n) = 4T\left(\frac{n}{4}\right) + n$

2. Second iteration: $T\left(\frac{n}{4}\right) = 4T\left(\frac{n}{16}\right) + \frac{n}{4}$ Substituting into the first equation: $T(n) = 4\left(4T\left(\frac{n}{16}\right) + \frac{n}{4}\right) + n = 16T\left(\frac{n}{16}\right) + 4n + n = 16T\left(\frac{n}{16}\right) + 5n$

3. Third iteration: $T\left(\frac{n}{16}\right) = 4T\left(\frac{n}{64}\right) + \frac{n}{16}$ Substituting into the previous expression: $T(n) = 16\left(4T\left(\frac{n}{64}\right) + \frac{n}{16}\right) + 5n = 64T\left(\frac{n}{64}\right) + 16n + 5n = 64T\left(\frac{n}{64}\right) + 21n$

Continuing this pattern, after $m$ iterations:

$T(n) = 4^m T\left(\frac{n}{4^m}\right) + n \left(\sum_{i=0}^{m-1} 4^i\right)$

Since $n = 4^k$ and $T(1) = 1$, we have $\frac{n}{4^k} = 1$ when $m = k$:

$T(n) = 4^k T(1) + n \left(\sum_{i=0}^{k-1} 4^i\right)$

$T(n) = 4^k \cdot 1 + n \left(\frac{4^k - 1}{3}\right)$

$T(n) = 4^k + \frac{n(4^k - 1)}{3}$

Since $n = 4^k$, $4^k = n$:

$T(n) = n + \frac{n(n - 1)}{3} = \frac{n^2 + 2n}{3}$

Thus, the explicit formula is:

$\boxed{T(n) = \frac{n^2 + 2n}{3}}$

Would you like more details on any of the steps? Here are some related questions you might find interesting:

1. How do these formulas relate to the Master Theorem for divide-and-conquer recurrences?
2. What are other common methods for solving recurrence relations?
3. How does the choice of base (3 in the first problem, 4 in the second) affect the solution?
4. What would the explicit formulas be for different initial conditions (e.g., $T(1) = 2$)?
5. How do these problems illustrate the growth rates of recursive functions?

Tip: Understanding the iterative method for solving recurrences is crucial for analyzing the time complexity of algorithms, especially in divide-and-conquer approaches.