## - Recurrence Relations and Mathematical Induction

A recurrence relation is an equation that defines a sequence based on a rule that gives the next term as a function of the previous term(s).

LESSON 1

A sequence is given by the following

\(u_1=4\)

\(u_{n+1}=u_n+3\)

Write down the first four terms of the sequence.

SOLUTION

\(u_1=4\)

\(u_2=u_{1+1}=u_1+3=7\)

\(u_3=u_{2+1}=u_2+3=10\)

\(u_4=u_{3+1}=u_3+3=13\)

MATHEMATICAL INDUCTION

The principle of Mathematical Induction has already been introduced in Unit 1 to prove summation and divisibility formulae. In this section we will extend this knowledge to sequences and inequalities.

LESSON 1

A sequence of positive integers is defined by

\(u_1=1\),

\(u_{n+1}=u_n+n(3n+1), n\in\mathbb{Z}^+\)

Prove by induction that \(u_n=n^2 (n-1)+1, n\in\mathbb{Z}^+\)

SOLUTION

\(u_n=n^2 (n-1)\)

- Show that \(u_1\) is true.

When \(n=1\),

\(u_1=1^2 (1-1)+1\)

\(u_1=1\)

Therefore \(u_1\) is true

- Assume that \(u_n\) is true for \(n=k\)

\(u_k=k^2 (k-1)+1\)

- Determine \(u_{k+1}\)

\(u_{k+1}=(k+1)^2 (k)+1\)

- Show that \(u_{k+1}\) is true \(∀ u_k\) is true

Now,

\(u_{k+1}=u_k+k(3k+1)\)

\(=k^2 (k-1)+1+k(3k+1)\)

\(=k^3-k^2+1+3k^2+k\)

\(=k^3+2k^2+k+1\)

\(=k(k^2+2k+1)+1\)

\(=k(k+1)^2+1\)

Therefore, \(u_{k+1}\) is true when \(u_k\) is true.

- Make concluding statement.

Hence, by Mathematical Induction

\(u_n=n^2 (n-1)+1, n\in \mathbb{Z}^+\)

LESSON 2

The sequence \(u_1, u_2, u_3, …\) where \(u_1=1\), is such that for \(n≥1\),

\(\displaystyle u_{n+1}={6u_n+4\over u_n+3}\)

By expressing \(4-u_{n+1}\) in terms of \(u_n\), prove by induction that \(u_n<4\) for \(n≥1\).

SOLUTION

Express \(4-u_{n+1}\) in terms of \(u_n\)

\(\displaystyle u_{n+1}={6u_n+4\over u_n+3}\)

\(\displaystyle 4-u_{n+1}=4-{6u_n+4\over u_n+3}\)

\(\displaystyle ={4(u_n+3)-(6u_n+4)\over u_n+3}\)

\(\displaystyle ={-2u_n+8\over u_n+3}\)

- State what you are trying to prove.

\(u_k<4\)

- Show that \(u_1\) is true.

\(u_1=1\)

\(∴u_1\) is true.

- Assume \(u_n\) is true for \(n=k\).

\(u_k<4\)

- Determine \(u_{k+1}\)

\(u_{k+1}<4 → 4-u_{k+1}>0\)

- Show that \(u_{k+1}\) is true \(∀ u_k\) is true.

\(\displaystyle 4-u_{n+1}={-2u_n+8\over u_n+3}\)

If

\(u_k=4\)

\(\displaystyle ⟹ 4-u_{k+1}={-2(4)+8\over 4+3}=0\)

However, since

\(u_k<4\)

\(⟹4-u_{k+1}>0\)

\(⟹4-u_{k+1}>0\)

\(⟹u_{k+1}<4\)

Therefore \(u_{k+1}\) is true \(∀ u_k\) is true.

- Make concluding statement.

Hence, by Mathematical Induction \(u_n<4\) for \(n≥1\).