BetterGrades Precalculus · Unit 15 · Lesson

Mathematical induction

Prove statements for all integers in a domain using base case and inductive step.

Textbook reading

The problem that opens the lesson

Prove 1+3+5+...+(2n1)=n21+3+5+...+(2n-1)=n^2.

Solution

Begin by identifying the mathematical object and the information that fixes it. State the proposition clearly, prove the base case, write the hypothesis, transform the k+1k+1 case using that hypothesis, and conclude for all permitted integers. The relevant conditions are not optional bookkeeping: Strong induction and multiple base cases are useful when a step depends on several earlier cases, but ordinary induction is the main spine here. Following that structure gives Verify n=1n=1; assume sum to kk is k2k^2; add 2k+12k+1 to obtain (k+1)2(k+1)^2.

Why this works

The method does not verify examples one by one; it proves a mechanism that carries truth forward indefinitely. The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Textbook reading

What this lesson is really about

Mathematical induction proves a statement for every integer in a sequence of allowed values.

The base case establishes the first link. The inductive hypothesis assumes one arbitrary case kk only for the purpose of proving the next case k+1k+1. Together, these create an unbroken logical chain.

The point is not merely to reproduce a formula. A learner should be able to identify the quantities or geometric objects involved, explain why the relationship has its stated form, and recognize when the same idea appears in a graph, table, diagram, or model.

Textbook reading

Why the relationship works

The method does not verify examples one by one; it proves a mechanism that carries truth forward indefinitely.

Textbook reading

A reliable way to work

State the proposition clearly, prove the base case, write the hypothesis, transform the k+1k+1 case using that hypothesis, and conclude for all permitted integers.

Strong induction and multiple base cases are useful when a step depends on several earlier cases, but ordinary induction is the main spine here.

After the symbolic work is complete, check the result. Depending on the lesson, this may mean substituting into an original equation, comparing coordinates, examining a graph, checking units, testing an interval, or confirming that every branch of a periodic solution has been included.

Textbook reading

What commonly goes wrong

A common error is assuming the k+1k+1 statement or failing to use the inductive hypothesis.

The repair is to return to the definition and identify the first step where the invalid solution stops describing the original mathematical object. Later algebra cannot rescue a first step that changed the domain, orientation, branch, or meaning of the problem.

Textbook reading

Worked examples

Worked example 1

Prove 1+3+5+...+(2n1)=n21+3+5+...+(2n-1)=n^2.

Solution

Begin by identifying the mathematical object and the information that fixes it. State the proposition clearly, prove the base case, write the hypothesis, transform the k+1k+1 case using that hypothesis, and conclude for all permitted integers. The relevant conditions are not optional bookkeeping: Strong induction and multiple base cases are useful when a step depends on several earlier cases, but ordinary induction is the main spine here. Following that structure gives Verify n=1n=1; assume sum to kk is k2k^2; add 2k+12k+1 to obtain (k+1)2(k+1)^2.

Why this works

The method does not verify examples one by one; it proves a mechanism that carries truth forward indefinitely. The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Transfer example

Problem

Prove an arithmetic-series identity.

Worked development

State the proposition clearly, prove the base case, write the hypothesis, transform the k+1k+1 case using that hypothesis, and conclude for all permitted integers. In this example, the first useful move is to make the defining structure visible rather than to search for a memorized answer. The base case establishes the first link. The inductive hypothesis assumes one arbitrary case kk only for the purpose of proving the next case k+1k+1. Together, these create an unbroken logical chain. Then apply the conditions explicitly: Strong induction and multiple base cases are useful when a step depends on several earlier cases, but ordinary induction is the main spine here. Finish by checking the result in a second representation and explaining what the result means.

Interpretation

Induction proves sum formulas, divisibility, inequalities, and recursive identities.

Reasoning example

Problem

Prove divisibility by induction.

Worked development

State the proposition clearly, prove the base case, write the hypothesis, transform the k+1k+1 case using that hypothesis, and conclude for all permitted integers. In this example, the first useful move is to make the defining structure visible rather than to search for a memorized answer. The base case establishes the first link. The inductive hypothesis assumes one arbitrary case kk only for the purpose of proving the next case k+1k+1. Together, these create an unbroken logical chain. Then apply the conditions explicitly: Strong induction and multiple base cases are useful when a step depends on several earlier cases, but ordinary induction is the main spine here. Finish by checking the result in a second representation and explaining what the result means.

Interpretation

Induction proves sum formulas, divisibility, inequalities, and recursive identities.

Worked example 4: quick check

What does the inductive hypothesis allow you to assume?

Solution

Begin by identifying the mathematical object and the information that fixes it. State the proposition clearly, prove the base case, write the hypothesis, transform the k+1k+1 case using that hypothesis, and conclude for all permitted integers. The relevant conditions are not optional bookkeeping: Strong induction and multiple base cases are useful when a step depends on several earlier cases, but ordinary induction is the main spine here. Following that structure gives The statement is true for one arbitrary allowed integer k, solely to prove the k+1k+1 case.

Why this works

The method does not verify examples one by one; it proves a mechanism that carries truth forward indefinitely. The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Domino-chain logic. Follow the foundation example from its given information to the conclusion. The labels identify the mathematical feature that makes the result valid: The method does not verify examples one by one; it proves a mechanism that carries truth forward indefinitely. The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.
Read this graph as text

Mathematical induction · Domino-chain logic. Follow the foundation example from its given information to the conclusion. The labels identify the mathematical feature that makes the result valid: The method does not verify examples one by one; it proves a mechanism that carries truth forward indefinitely. The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion. The figure uses concrete points, curves, arrows, intervals, or matrix structure instead of relying on color alone.

Labels, point shapes, line styles, arrows, and position carry the mathematical meaning; color is supplementary.

Why it matters: Use the mathematical objects in this figure to support the lesson outcome: Prove statements for all integers in a domain using base case and inductive step.

Anchor figure · Domino-chain logic

Follow the foundation example from its given information to the conclusion. The labels identify the mathematical feature that makes the result valid: The method does not verify examples one by one; it proves a mechanism that carries truth forward indefinitely. The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Base-hypothesis-step template. Read the numbered reasoning path in order. Each stage preserves the quantities, restrictions, or structural conditions needed for mathematical induction.
Read this graph as text

Mathematical induction · Base-hypothesis-step template. Read the numbered reasoning path in order. Each stage preserves the quantities, restrictions, or structural conditions needed for mathematical induction. The figure uses concrete points, curves, arrows, intervals, or matrix structure instead of relying on color alone.

Labels, point shapes, line styles, arrows, and position carry the mathematical meaning; color is supplementary.

Why it matters: Use the mathematical objects in this figure to support the lesson outcome: Prove statements for all integers in a domain using base case and inductive step.

Mechanism figure · Base-hypothesis-step template

Read the numbered reasoning path in order. Each stage preserves the quantities, restrictions, or structural conditions needed for mathematical induction.

Invalid circular induction comparison. Compare the valid path with the tempting shortcut. The figure shows why assuming the k+1 statement or failing to use the inductive hypothesis leads to a false conclusion.
Read this graph as text

Mathematical induction · Invalid circular induction comparison. Compare the valid path with the tempting shortcut. The figure shows why assuming the k+1 statement or failing to use the inductive hypothesis leads to a false conclusion. The figure uses concrete points, curves, arrows, intervals, or matrix structure instead of relying on color alone.

Labels, point shapes, line styles, arrows, and position carry the mathematical meaning; color is supplementary.

Why it matters: Use the mathematical objects in this figure to support the lesson outcome: Prove statements for all integers in a domain using base case and inductive step.

Comparison and error figure · Invalid circular induction comparison

Compare the valid path with the tempting shortcut. The figure shows why assuming the k+1k+1 statement or failing to use the inductive hypothesis leads to a false conclusion.

Textbook reading

Application and interpretation

Induction proves sum formulas, divisibility, inequalities, and recursive identities.

A contextual answer must include units, a meaningful domain, and the assumptions that make the model plausible. An exact mathematical relationship should not be diluted into a decimal unless a measurement or comparison requires it.

Check yourself

What does the inductive hypothesis allow you to assume?

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Practice

Ten concrete questions

Practice 101

What does the inductive hypothesis allow you to assume?

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Practice 202

Prove an arithmetic-series identity.

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Practice 303

Prove divisibility by induction.

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Practice 404

Diagnose a proof that assumes the k+1k+1 case.

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Practice 505

State the defining idea behind mathematical induction in one precise sentence.

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Practice 606

What condition or domain restriction must remain visible in the solution?

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Practice 707

Describe the most likely incorrect first step and explain why it fails.

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Practice 808

Translate the main result into a second representation: graph, diagram, table, equation, or context.

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Practice 909

Write one exact conclusion and one corresponding numerical approximation or verbal interpretation.

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Practice 1010

Explain how this lesson's idea will be used later in the course.

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Textbook reading

Lesson summary

Mathematical induction proves a statement for every integer in a sequence of allowed values.

The central condition to remember is this: Strong induction and multiple base cases are useful when a step depends on several earlier cases, but ordinary induction is the main spine here.

Connection forward

The next lesson studies binomial coefficients and their recursive pattern.

The next lesson is Pascal's triangle and binomial coefficients.

Source record

Original BetterGrades manuscript, rights-separated references.

  • Stitz & Zeager, Precalculus, Chapter 9
  • University of Washington Precalculus, discrete-model problems
  • AP Precalculus framework, sequence and model connections

No long source passage is reproduced.