- 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 \(n≥1\).