Question

The terms of a sequence are given recursively as \(a_{0}=0, a_{1}=4,\) and \(a_{n}=8 a_{n-1}-\) \(16 a_{n-2}\) for \(n \geq 2\). Write out the information that the inductive step assumes and what the step must prove in proving \(b_{n}=n 4^{n}\) is a closed form for the sequence. Suppose \(n_{0}=1\) and the base cases are 0 and \(1 .\)

Short answer

The closed form \(b_n = n \cdot 4^n\) is proven for the sequence using induction.

Step-by-step solution

Understand the Recursive Formula

The sequence is defined recursively with the initial conditions \(a_{0} = 0\) and \(a_{1} = 4\). The recursive relation for \(n \geq 2\) is given by \(a_{n} = 8a_{n-1} - 16a_{n-2}\). Our goal is to show that this sequence can also be represented in the closed form \(b_n = n \cdot 4^n\).

Base Cases Verification

Check the base cases for the assumed closed form. For \(n = 0\), \(b_0 = 0 \cdot 4^0 = 0\), which matches \(a_0 = 0\). For \(n = 1\), \(b_1 = 1 \cdot 4^1 = 4\), which matches \(a_1 = 4\). Both base cases are satisfied.

Formulate the Inductive Hypothesis

Assume for some \(k \geq 1\) that the closed form holds for \(n = k\) and \(n = k-1\), i.e., \(a_k = k \cdot 4^k\) and \(a_{k-1} = (k-1) \cdot 4^{k-1}\). This assumption will be used to prove the closed form for \(n = k+1\).

Apply the Inductive Step

Use the recursive formula to express \(a_{k+1}\):\[a_{k+1} = 8a_k - 16a_{k-1}.\] Substituting the inductive hypothesis, \(a_k = k \cdot 4^k\) and \(a_{k-1} = (k-1) \cdot 4^{k-1}\), into the equation, we get: \[a_{k+1} = 8(k \cdot 4^k) - 16((k-1) \cdot 4^{k-1}).\]

Simplify the Inductive Step

Simplify the expression for \(a_{k+1}\) using the properties of exponents and algebra: \[a_{k+1} = 8k \cdot 4^k - 16(k-1) \cdot 4^{k-1}\] becomes \[a_{k+1} = 8k \cdot 4^k - 16k \cdot 4^{k-1} + 16 \cdot 4^{k-1}.\] Factor out \(4^{k-1}\): \[a_{k+1} = (8k \cdot 4 - 16k + 16)4^{k-1}.\] Simplifying inside the parentheses, \[a_{k+1} = (32k - 16k + 16)4^{k-1} = ((16k + 16)4^{k-1})\] factors to \[a_{k+1} = (16(k + 1))4^{k-1} = (k+1)\cdot4^{k+1}.\] Thus, \(b_{k+1} = (k+1) \cdot 4^{k+1}\) matches \(a_{k+1}\).

Conclusion

The inductive step is verified, proving that if the formula holds for \(n = k\), it will also hold for \(n = k+1\). Since the base cases are satisfied, by the principle of induction, the closed form \(b_n = n \cdot 4^n\) is valid for all \(n \geq 0\).

Key concepts

Recursive Sequences

Understanding recursive sequences is key to solving many mathematical and programming problems. A recursive sequence is defined by relating each term to one or more of its predecessors. This form of defining a sequence allows us to generate terms step-by-step using initial values and a recurrence relation.

In the exercise, the sequence starts with base terms:

• For the zero term, we have: \(a_0 = 0\)
• For the first term, we have: \(a_1 = 4\)

Then, for all subsequent terms where \(n \geq 2\), they are defined by the relation \(a_n = 8a_{n-1} - 16a_{n-2}\). This means each term relies on the two terms before it.
Recursive sequences are like building a tower block by block, using the few known blocks at the base to create further blocks above.

Closed Form Solution

A closed form solution provides a way to directly compute any term in a sequence without having to rely on previous terms. This is highly useful as it allows us to calculate, for instance, the 1000th term without calculating all the previous ones.

For our sequence, the closed form solution is given as \(b_n = n \cdot 4^n\). This formula gives the result for any term \(a_n\) as a function of \(n\), skipping the recursive calculations.
Such closed forms are generally more efficient and reveal deeper understanding of the sequence structure. It sheds light on patterns that are not immediately apparent in the recursive definition.

Inductive Hypothesis

The inductive hypothesis is a crucial part of proving statements about sequences, especially when proving they follow a certain closed form pattern. It involves assuming the statement is true for a general term \(k\), and often for \(k-1\) too, and then using this assumption to prove it for the next term, \(k+1\).

In this exercise, the inductive hypothesis involves assuming:

• \(a_k = k \cdot 4^k\)
• \(a_{k-1} = (k-1) \cdot 4^{k-1}\)

By accepting these assumptions, we aim to show that the formula holds true for \(a_{k+1}\) as well. The hypothesis forms the backbone of the proof, bridging known values and extending them.

Base Case Verification

Base case verification is often the first step in proving by induction. It involves checking initial conditions or starting points to ensure they satisfy the formula or proposition to be proven.

In our specific case, two base cases were checked:

• For \(n=0\), we have \(b_0 = 0 \cdot 4^0 = 0\), matching \(a_0 = 0\).
• For \(n=1\), we have \(b_1 = 1 \cdot 4^1 = 4\), matching \(a_1 = 4\).

These verifications ensure that at the outset of the sequence, the closed form matches the recursive results. Validating these cases guarantees that our base is stable, providing a solid foundation to build upon further with the inductive step.