- 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 $$n≥1$$.