Solving the Recurrence Relation fn-1 3fn for f3

In this article, we will explore the solution to the recurrence relation fn-1 3fn, given the initial condition f0 4. We will walk through the method to determine f3, using a step-by-step approach and validating our solution through mathematical induction.

Step-by-Step Solution

Given the recurrence relation fn-1 3fn, we can express each term in the sequence in terms of the previous term. Let's start with the initial term f0 4.

Step 1: Derived Values

For n 1: f1 3f0 3(4) 12

For n 2: f2 3f1 3(12) 36

For n 3: f3 3f2 3(36) 108

Thus, the value of f3 is 108.

Second Method: Geometric Sequence

The recurrence relation can also be viewed as a geometric sequence where the common ratio is 3 and the first term is 4. The general form of the geometric sequence is:

fn 3n cdot 4

Using this formula, we can find:

For n 3,

f3 33 cdot 4 27 cdot 4 108

Third Method: Induction Proof

We can prove the above result using mathematical induction:

Base Case

f0 4 30 cdot 4

Base case holds true.

Induction Hypothesis

Assume fn 3n cdot 4 for some n.

Induction Step

For n 1: fn 1 3fn 3(3n cdot 4) 3^{n 1} cdot 4

Hence, by induction, the general term is fn 3n cdot 4, and thus f3 33 cdot 4 108.

Fourth Method: Difference Equation Approach

The recurrence relation fn-1 3fn can be solved using the shift operator [E - 3] 0, where E represents the shift operator. The characteristic equation is m - 3 0, with m 3. Therefore, the general form of the solution is f(n) A.3^n, where A is a constant determined by the initial condition.

Using the initial condition f0 4, we get:

A.3^0 4 4

Hence, A 4. Thus, the solution is:

f(n) 4.3^n

Therefore, f3 4.3^3 108.

Conclusion

Through several methods, we have shown that for the given recurrence relation fn-1 3fn and initial condition f0 4, the value of f3 is 108. This exploration showcases the versatility of mathematical techniques in solving recurrence relations and highlights the importance of both pattern recognition and rigorous proof.