Constructing the induction step | Year 12

01 / LEARN

The idea, step by step

Expose what changes

A sum ending at k+1 equals the sum ending at k plus one new term. The new square is (k+1)^2, and the new odd number is 2k+1. Write that bridge before using the hypothesis.

For a product, multiply by the new factor instead. The target at k+1 guides the algebra but cannot be assumed as an extra premise.

Reach the target legally

Replace the old expression by its assumed value, then simplify with exact algebra. Factor a common k+1 where this makes the target clear. Substitute k+1 everywhere in the original formula to identify the required endpoint.

Finish by stating that P(k+1) follows from P(k). The conclusion about every integer needs both this implication and the base case.

A common trap

Using P(k+1) to prove itself.

Instead: Use only P(k), algebra and previously known facts.

02 / WORKED EXAMPLES

See the thinking

Example 1

An odd-number total

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

At $n = 1$ , $1 = 1^{2}$ .
Assume the first $k$ odd numbers sum to $k^{2}$ .
Add the next odd term $2 k + 1$ .
The result is $k^{2} + 2 k + 1 = {(k + 1)}^{2}$ .
Conclude all positive integers by induction.

1 + 3 + \dots + (2 n - 1) = n^{2}

is true throughout the stated domain.

Example 2

Extending a product

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

At $n = 1$ , the product is $2$ .
Assume the product to $k$ is $k + 1$ .
Multiply by $(k + 2) / (k + 1)$ .
The product becomes $k + 2$ .
Conclude all positive integers by induction.

\prod_{r = 1}^{n} (r + 1) / r = n + 1

is true throughout the stated domain.

Check your understanding

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

Show answer

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

Replace r with k+1 in both factors.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Simplify $k^{2} + 2 k + 1$ and explain its induction use.
2 marks
Given $S_{k} = k (k + 1) / 2$ , 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 $\prod_{r = 1}^{k} (r + 1) / r = k + 1$ , 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 $S_{k} = k (k + 1) / 2$ , derive $S_{k + 1}$ when the next term is k+1.
3 marks
$S_{k + 1} = S_{k} + (k + 1)$ .
Substitute and factor $(k + 1) (k / 2 + 1)$ .
$S_{k + 1} = (k + 1) (k + 2) / 2$ .
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 $\prod_{r = 1}^{k} (r + 1) / r = k + 1$ , find the next product.
3 marks
Multiply by $(k + 2) / (k + 1)$ .
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 ↗

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