What is weak algebra in the step?

For a sum, you must factor to the exact target form; leaving an unsimplified expression loses the accuracy marks.

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How does proof by mathematical induction work, and what must each step contain to be rigorous?

Proof by mathematical induction for summation formulae, divisibility results, recurrence relations and powers of matrices, with a correctly stated base case, inductive hypothesis, inductive step and conclusion.

A focused answer to the OCR A-Level Further Mathematics A content on proof by mathematical induction, covering the structure (base case, inductive hypothesis, inductive step and conclusion) and its use for summation formulae, divisibility results, recurrence relations and powers of a matrix, with the rigorous wording examiners require.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The structure of an inductive proof
Induction for summation formulae
Induction for divisibility
Induction for recurrence relations and matrix powers
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What this dot point is asking

OCR wants you to prove statements by mathematical induction with full rigour: state and verify a base case, assume the statement for $n = k$ (the inductive hypothesis), prove it follows for $n = k + 1$ (the inductive step), and write a complete concluding statement. The standard contexts are summation formulae, divisibility results, recurrence relations and powers of a matrix.

The structure of an inductive proof

Induction works like dominoes: the base case knocks over the first, and the inductive step guarantees that each domino knocks over the next, so all fall. Every part must be present and the conclusion explicitly stated, because OCR awards a mark for the full concluding sentence.

Induction for summation formulae

For a sum, the inductive step adds the next term to the assumed result, then rearranges to match the formula with $n = k + 1$ . The skill is the algebraic factorisation that produces the target form.

Induction for divisibility

For a divisibility result, write the inductive hypothesis as "the expression equals a multiple of the divisor", then manipulate the $n = k + 1$ expression to reveal a factor of the divisor. A common trick is to add and subtract a term to introduce the assumed multiple.

Induction for recurrence relations and matrix powers

For a recurrence-defined sequence, assume the closed form for $u_k$ (and sometimes $u_{k-1}$ ) and use the recurrence to obtain $u_{k+1}$ . For a power of a matrix, assume the form of $\mathbf{M}^k$ and multiply by $\mathbf{M}$ to obtain $\mathbf{M}^{k+1}$ , showing it matches the conjectured pattern.

Prove a matrix power

Prove that $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ for all positive integers $n$ .

step 1 Base case

$n = 1$ : the right side is $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ , which matches.

step 2 Inductive step

Assume $\mathbf{M}^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ . Then $\mathbf{M}^{k+1} = \mathbf{M}^k\mathbf{M} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & k+1 \\ 0 & 1 \end{pmatrix}$ .

step 3 Conclude

This matches the formula with $n = k+1$ , so by induction it holds for all positive integers $n$ .

Induction is the rigorous backbone of the series strand and is used to justify the summation formulae and de Moivre's theorem.

Try this

Q1. State the four parts of an inductive proof. [2 marks]

Cue. Base case, inductive hypothesis, inductive step, conclusion.

Q2. In proving $5^n - 1$ is divisible by $4$ , how do you write the inductive hypothesis? [1 mark]

Cue. $5^k - 1 = 4m$ for some integer $m$ , that is $5^k = 4m + 1$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20196 marksProve by induction that

\displaystyle\sum_{r=1}^{n} r(r+1) = \dfrac{1}{3}n(n+1)(n+2)

for all positive integers

n

Show worked answer →

Base case $n = 1$ (B1): LHS $= 1 \times 2 = 2$ ; RHS $= \tfrac{1}{3}(1)(2)(3) = 2$ . True for $n = 1$ .

Assume true for $n = k$ (M1): $\displaystyle\sum_{r=1}^{k} r(r+1) = \tfrac{1}{3}k(k+1)(k+2)$ .

For $n = k+1$ , add the next term (M1): $\displaystyle\sum_{r=1}^{k+1} r(r+1) = \tfrac{1}{3}k(k+1)(k+2) + (k+1)(k+2)$ .

Factor out $(k+1)(k+2)$ (A1): $= (k+1)(k+2)\left(\tfrac{1}{3}k + 1\right) = (k+1)(k+2)\cdot\tfrac{k+3}{3} = \tfrac{1}{3}(k+1)(k+2)(k+3)$ (A1).

This is the formula with $n = k+1$ , so if true for $k$ it is true for $k+1$ ; true for $n = 1$ , hence true for all positive integers $n$ by induction (A1).

Markers reward the base case, the assumption, adding the next term, factorising to the target form, and a complete concluding statement.

OCR 20226 marksProve by induction that

7^n - 1

is divisible by

6

for all positive integers

n

Show worked answer →

Base case $n = 1$ (B1): $7^1 - 1 = 6$ , which is divisible by $6$ .

Assume true for $n = k$ (M1): $7^k - 1 = 6m$ for some integer $m$ , so $7^k = 6m + 1$ .

Consider $n = k+1$ (M1): $7^{k+1} - 1 = 7 \cdot 7^k - 1 = 7(6m + 1) - 1 = 42m + 7 - 1 = 42m + 6$ (A1).

Factor (A1): $= 6(7m + 1)$ , which is divisible by $6$ .

So if true for $k$ it is true for $k+1$ ; true for $n = 1$ , hence true for all positive integers $n$ by induction (A1).

Markers reward the base case, the assumption written as a multiple of $6$ , the substitution $7^k = 6m + 1$ , the factor of $6$ , and the conclusion.

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Sources & how we know this

OCR A Level Further Mathematics A (H245) specification — OCR (2017)