(5.7.17) Solving a Recurrence Relation Using Iteration to Derive an Explicit Formula

In this presentation, I will walk you through the process of solving a recurrence relation using the method of iteration to derive an explicit formula for the sequence. We start with the recurrence relation defined as T_n = 2T_(n-1) + 3, where T_1 equals 3. My preference for iteration comes into play here as I find it to be a clear and effective method for solving such relations.

To begin, we calculate T_2 in terms of T_1. According to the recurrence relation, T_2 equals 2T_1 + 3. I will leave this expression as is for now without substituting T_1's value until we finalize our generalization. Next, we look for T_3 using the recurrence relation, leading to T_3 being expressed as 2T_2 + 3. At this point, we will substitute our earlier expression for T_2 into this equation, resulting in T_3 = 2(2T_1 + 3) + 3. Rather than performing the multiplication just yet, I will keep it in a factored form, writing it as 2^2T_1 + 2*3 + 3.

Continuing this pattern, we derive T_4 using T_3, resulting in T_4 = 2T_3 + 3. Again substituting our earlier result for T_3 yields T_4 = 2(2^2T_1 + 2*3 + 3) + 3. This simplification reveals a pattern emerging, where T_4 can be expressed in terms of powers of 2. As I predict T_5, it follows a similar structure, suggesting a general formula for T_n.

Eventually, we reach a generalization for T_n, which I express in summation form. The explicit formula showcases the contributions from the geometric series formed by the powers of 2 and their associated coefficients. This allows us to compute terms like T_100 without needing to evaluate all preceding terms, significantly simplifying our calculations.

I appreciate your attention, and I look forward to exploring more mathematical concepts in our next session.

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