Constructing the induction step | Year 12

01 / LEARN

Understand the idea.

Expose what changes

The induction step must connect the new expression to the case you are allowed to assume. For a sum, separate its new term; for a product, separate its new factor. The preceding terms or factors then match the hypothesis exactly.

Substitute $r = k + 1$ into the entire term. Thus $2 r - 1$ becomes $2 k + 1$ , while $r^{2}$ becomes ${(k + 1)}^{2}$ . Writing this bridge first prevents an unnoticed change to the old part.

Reach the target legally

Write the required formula at $k + 1$ separately as a target. Work from the new sum or product, replace only the old part using the hypothesis, and simplify until you reach that target.

The target guides your factorisation; it is not an extra assumption. After the algebra, state that the arbitrary $k$ case implies the next case. A checked base then establishes the whole positive-integer domain.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An odd-number total

Prove $1 + 3 + \dots + (2 n - 1) = n^{2}$ for $n \geq 1$ .

Worked solution

1
Verify the starting case.
$n = 1 : 1 = 1^{2} .$
The first positive integer satisfies the identity.
2
Assume one arbitrary case.
$\sum_{r = 1}^{k} (2 r - 1) = k^{2}, k \in ℤ, k \geq 1.$
This is the temporary hypothesis, not the conclusion for every integer.
3
Separate the next odd number.
$\sum_{r = 1}^{k + 1} (2 r - 1) = \sum_{r = 1}^{k} (2 r - 1) + (2 k + 1) .$
Substituting the new index into the term gives $2 (k + 1) - 1$ .
4
Use the hypothesis and complete the square.
$\begin{aligned} \sum_{r = 1}^{k + 1} (2 r - 1) & = k^{2} + 2 k + 1 \\ = {(k + 1)}^{2} . \end{aligned}$
The new total has exactly the formula required at $k + 1$ .
5
Conclude by induction.
$\sum_{r = 1}^{n} (2 r - 1) = n^{2} (n \in ℤ, n \geq 1) .$
The true base and the successor implication establish every claimed case.

Answer

$1 + 3 + \dots + (2 n - 1) = n^{2}$ is true throughout the stated domain.

EXAMPLE 02

Extending a product

Prove $\prod_{r = 1}^{n} (r + 1) / r = n + 1$ for $n \geq 1$ .

Worked solution

1
Check the first product.
$n = 1 : \prod_{r = 1}^{1} \frac{r + 1}{r} = \frac{2}{1} = 2 = 1 + 1.$
The single-factor product agrees with the proposed formula.
2
State the hypothesis.
$\prod_{r = 1}^{k} \frac{r + 1}{r} = k + 1, k \in ℤ, k \geq 1.$
Assume precisely the product of the first $k$ factors.
3
Multiply by the new factor.
$\prod_{r = 1}^{k + 1} \frac{r + 1}{r} = (\prod_{r = 1}^{k} \frac{r + 1}{r}) \frac{k + 2}{k + 1} .$
A product is extended by multiplication, with $r = k + 1$ in the new factor.
4
Substitute and cancel.
$\begin{aligned} \prod_{r = 1}^{k + 1} \frac{r + 1}{r} & = (k + 1) \frac{k + 2}{k + 1} \\ = k + 2 = (k + 1) + 1. \end{aligned}$
The factor $k + 1$ is nonzero because $k \geq 1$ .
5
State the full result.
$\prod_{r = 1}^{n} \frac{r + 1}{r} = n + 1 (n \in ℤ, n \geq 1) .$
Mathematical induction combines the base with the proved next-case implication.

Answer

$\prod_{r = 1}^{n} (r + 1) / r = n + 1$ is true throughout the stated domain.

03 / YOUR TURN

Try it yourself.

Check your understanding

What is the next term after $\sum_{r = 1}^{k} r (r + 1)$ ?

Show answer & reasoning

$(k + 1) (k + 2)$ .

The new index is $r = k + 1$ , so the next term is $(k + 1) ((k + 1) + 1) = (k + 1) (k + 2)$ . Both factors must change.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Simplify $k^{2} + 2 k + 1$ and explain its induction use.
2 marks
Given $S_{k} = \frac{k (k + 1)}{2}$ , derive $S_{k + 1}$ when the next term is $k + 1$ .
3 marks
What term extends $\sum_{r = 1}^{k} r^{2}$ to the sum ending at $k + 1$ ?
1 mark
If $\prod_{r = 1}^{k} \frac{r + 1}{r} = k + 1$ , find the next product.
3 marks
A student writes the target at $k + 1$ immediately after the hypothesis at $k$ . What is missing?
2 marks
For $A_{n} = 5^{n} - 1$ , establish $A_{k + 1} = 5 A_{k} + 4$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Simplify $k^{2} + 2 k + 1$ and explain its induction use.
2 marks
The new odd term is $2 k + 1$ .
Add it to the assumed odd-number sum $k^{2}$ .
${(k + 1)}^{2}$ .
Given $S_{k} = \frac{k (k + 1)}{2}$ , derive $S_{k + 1}$ when the next term is $k + 1$ .
3 marks
$S_{k + 1} = S_{k} + (k + 1)$ .
Substitute and factor $(k + 1) (\frac{k}{2} + 1)$ .
$S_{k + 1} = \frac{(k + 1) (k + 2)}{2}$ .
What term extends $\sum_{r = 1}^{k} r^{2}$ to the sum ending at $k + 1$ ?
1 mark
Substitute the new index into the entire term.
${(k + 1)}^{2}$ .
If $\prod_{r = 1}^{k} \frac{r + 1}{r} = k + 1$ , find the next product.
3 marks
Multiply by $\frac{k + 2}{k + 1}$ .
Cancel the nonzero common factor.
$k + 2$ .
A student writes the target at $k + 1$ immediately after the hypothesis at $k$ . What is missing?
2 marks
For a sum, expose the next term.
Use the hypothesis only for the old sum.
A justified bridge from the new expression to the old expression.
For $A_{n} = 5^{n} - 1$ , establish $A_{k + 1} = 5 A_{k} + 4$ .
3 marks
Expanding gives $5^{k + 1} - 5 + 4$ .
The identity connects the new case to the old case.
$5^{k + 1} - 1 = 5 (5^{k} - 1) + 4$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist mathematical induction. Explicit summation, power and divisibility identities; induction of inequalities or recursive formulae is not required.

Official curriculum source ↗

In this lesson

Understand the idea.

Expose what changes

Reach the target legally

See the method in action.

An odd-number total

Verify the starting case.

Assume one arbitrary case.

Separate the next odd number.

Use the hypothesis and complete the square.

Conclude by induction.

Extending a product

Check the first product.

State the hypothesis.

Multiply by the new factor.

Substitute and cancel.

State the full result.

Try it yourself.

Check your understanding

Keep practising