Question

The terms of a sequence are given recursively as \(p_{0}=1, p_{1}=2,\) and \(p_{n}=2 p_{n-1}-\) \(p_{n-2}\) for \(n \geq 2\). Prove by induction that \(b_{n}=1+n\) is a closed form for the sequence.

Short answer

The closed form \(b_n = 1+n\) is valid for all \(n \geq 0\) by induction.

Step-by-step solution

Base Case Verification

First, verify that the closed form works for the initial term of the sequence. Let's try \(n=0\) and \(n=1\). For \(n=0\), the closed form \(b_n = 1+n\) gives us \(b_0 = 1+0 = 1\). The recursive sequence gives \(p_0 = 1\), so the closed form matches the initial term. For \(n=1\), \(b_1 = 1+1 = 2\), and from the recursive formula, \(p_1 = 2\). Hence, the closed form works for the initial terms.

Inductive Hypothesis

Assume that the closed form \(b_n = 1+n\) works for a general term \(n=k\). That is, assume \(b_k = 1+k = p_k\). Also assume it works for \(n=k-1\), so \(b_{k-1} = 1+(k-1) = k = p_{k-1}\). We aim to show it holds for \(n=k+1\).

Inductive Step Verification

Apply the recursive relation to show the closed form works for \(n = k+1\). According to the recursive formula \(p_{k+1} = 2p_k - p_{k-1}\). Substitute \(p_k = 1+k\) and \(p_{k-1} = k\) (as per the inductive hypothesis) into the relation: \[p_{k+1} = 2(1+k) - k = 2+2k-k = 1+k+1\].This matches the closed form \(b_{k+1} = 1 + (k+1)\). Hence, the closed form holds for \(n = k+1\).

Conclusion of Induction

Since the closed form \(b_n = 1+n\) holds true for the initial cases \(n = 0\) and \(n = 1\), and the inductive step shows that if it holds for \(n = k\), it also holds for \(n = k+1\), by mathematical induction, the closed form is valid for all \(n \geq 0\).

Key concepts

Recursive Sequences

A recursive sequence is essentially a sequence of numbers where each term is defined in terms of one or more previous terms. This means that you don't just get to calculate each number from scratch; you rely on the numbers that came before it. This is why it is called "recursive," from the word "recursion," meaning "going back."

In the problem we are examining, the sequence is given recursively by:

• The pre-defined first term, called the initial condition, is: \( p_0 = 1 \).


• The second term is: \( p_1 = 2 \).
• Then for any term \( p_n \), when \( n \) is greater than or equal to 2, it is produced using the rule: \( p_n = 2p_{n-1} - p_{n-2} \).

What makes recursive sequences interesting is that they often reflect the very nature of a problem, like how certain patterns build on themselves. Understanding these fundamentals helps you comprehend how sequences evolve over time, and honing this skill becomes vital when you deal with more complex problems down the line. This understanding also sets the stage for forming and proving closed forms, which we will discuss next.

Closed Form

The term 'closed form' refers to a way of expressing a sequence without needing to reference previous terms. Think of it as having a magical shortcut that can take you to any term in the sequence directly without stepping through every preceding one.

For our sequence, the closed form is stated as \( b_n = 1 + n \). What this means is, for any value of \( n \), we can compute the term by simply adding 1 to \( n \). This is indeed a more straightforward and intuitive route, avoiding the iterative steps of a recursive sequence.

The beauty of having a closed form is multi-fold:

• It simplifies and reduces the effort needed to find specific terms in the sequence.


• It offers insight into the overall behavior of the sequence, helping you identify patterns or systematic flaws in your set of numbers.


• Closed forms can also significantly reduce computational time, especially in large-scale calculations.

For proving that \( b_n = 1 + n \) truly represents the sequence, we rely on mathematical induction—one clever tool in our mathematical toolbox, which we touch on next.

Inductive Hypothesis

When trying to prove statements, especially those regarding sequences, the inductive hypothesis plays a crucial role in the process of mathematical induction. This method allows us to prove a property for every natural number by showing it holds true for a given starting point and assuming it holds for a point \( n \), then proving it for the next point \( n+1 \).

The inductive hypothesis in this context is the assumption that if our statement, which is \( b_n = 1 + n \), holds for \( n = k \) and \( n = k-1 \), it will also hold for \( n = k+1 \). Essentially, you hypothesize that the closed form correctly produces the sequence for an arbitrary step in the sequence \( n = k \).

• You start by proving it is true for the base cases, often \( n=0 \) and \( n=1 \).
• You assume that the formula works for some \( n=k \), and under this assumption, you demonstrate it works for \( n=k+1 \).

This step is crucial! It is where the power of induction shines. By verifying it for \( n+1 \) based on an assumption, you extend the truth of the statement to all natural numbers that follow. When you are done, you've proven with certainty that your closed form correctly expresses the sequence across the board.