91Ó°ÊÓ

In recurrence relations, identifying the base case is crucial. It's a predefined value that allows us to start the iterative process. For this exercise, the base case given is \( f(1) = 1 \).
The base case provides a starting point to begin solving the recurrence relation. Once we have the base case, we can use it to apply the recurrence relation for larger values of \( n \). In our example, we know that for \( n = 1 \), the function value is \( f(1) = 1 \), which sets a clear reference point.

Iterative substitution involves repeatedly applying the recurrence relation to break down the problem. Let's walk through the iterative steps of our example:
For \( n = 4 \, f(4) \) can be obtained by substituting \( n = 4 \) into the recurrence relation:

• \( f(4) = 5 f(1) + 6 \times 4 \)
• \( f(4) = 5 \times 1 + 24 \)
• \( f(4) = 29 \)

Next, for \( n = 4^{2} = 16 \, f(16) \):

• \( f(16) = 5 f(4) + 6 \times 16 \)
• \( f(16) = 5 \times 29 + 96 \)
• \( f(16) = 241 \)

Finally, for \( n = 4^{3} = 64 \, f(64) \):

• \( f(64) = 5 f(16) + 6 \times 64 \)
• \( f(64) = 5 \times 241 + 384 \)
• \( f(64) = 1589 \)

By iterative substitution, we can see the values that our recurrence relation generates. This helps in recognizing any patterns or regularities, which can lead to a general formula.

The geometric series concept is integral in solving recurrence relations, especially when we seek the pattern and general formula. In our problem, the geometric series helps sum the components:
After identifying the iterative pattern, we realize it forms an additive series:
\( 4^k + 4^{k-1} + \frac{4^{k}}{4} + \frac{4^{0}}{4^{1}} \). This is a geometric series with the first term \( a = 4^{k} \) and common ratio \( r = \frac{1}{4} \).
The sum of the infinite geometric series is given as: \[ S = \frac{a}{1-r} \]
In our context:

• \( S = 4^k \times \frac{1}{1 - \frac{1}{4}} \)
• \( S = 4^k \times \frac{4}{3} \)
• \( S = \frac{4^{k+1}}{3} \)

Thus, our recurrence relation simplifies using the geometric series sum. We derive:
\[ f(4^k) = 5^k + 6 \times \frac{4^{k+1} -1}{3} = 5^k + 8 \times (4^k - 1) \]. This final expression shows how geometric series parts contribute to the total, providing a closed-form solution.