Skip to content

Edexcel A-level Further Maths revision notes

Proof

Section 1
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
1 specification point

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section 1

Checked against Edexcel 9FM0 section 1. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

In the exam: Formulae booklet provided · calculator allowed in every paper

Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Construct proofs using mathematical induction. Contexts include sums of series, divisibility and powers of matrices.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Mathematical induction proves a statement P(n)P(n) for every integer from a stated starting value. First verify the base case.
  • Then assume P(k)P(k) is true for an arbitrary permitted integer kk; this is the induction hypothesis. Use that hypothesis to derive P(k+1)P(k+1), rather than assuming the next case.
  • For a series, add the term with index k+1k+1; for divisibility, expose the required integer factor; for a matrix power, multiply the assumed expression for AkA^k by AA.
  • The conclusion must state that the base case and inductive implication together prove P(n)P(n) for all integers in the stated range.
  • Checking several cases alone is not a proof.
The four linked stages of a proof by mathematical induction.
Worked example

Prove by mathematical induction that 52n−15^{2n}-1 is divisible by 2424 for every positive integer nn.

  1. 1.For n=1n=1, 52−1=245^2-1=24, so the base case is divisible by 2424.
  2. 2.Assume 52k−1=24m5^{2k}-1=24m for some positive integer kk and some integer mm.
  3. 3.52(k+1)−1=25(52k−1)+245^{2(k+1)}-1=25\left(5^{2k}-1\right)+24.
  4. 4.Substitution gives 52(k+1)−1=24(25m+1)5^{2(k+1)}-1=24(25m+1), which is divisible by 2424 because 25m+125m+1 is an integer.
  5. 5.Therefore 52n−15^{2n}-1 is divisible by 2424 for every positive integer nn.

Answer: The base case holds and P(k)⇒P(k+1)P(k)\Rightarrow P(k+1), so the divisibility result follows by mathematical induction.

Common mistakes

  • Don't fall into the trap of assuming the statement for k+1k+1, which makes the inductive step circular.
  • Don't fall into the trap of writing only that the result is a multiple of 2424 without expressing the remaining factor as an integer.
  • Don't fall into the trap of checking several numerical values and treating those examples as a proof for all positive integers.

Exam tip

A proof-by-induction response must show the base case, use the named induction hypothesis in the k+1k+1 case, and finish with a quantified conclusion.

Tier 1 · Easy

ORIGINAL

1.

Prove by mathematical induction that 1+3+5+⋯+(2n−1)=n21+3+5+\cdots+(2n-1)=n^2 for every positive integer nn.

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

Prove by mathematical induction that 8n−18^n-1 is divisible by 77 for every positive integer nn.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

Let A=(2102)A=\begin{pmatrix}2&1\\0&2\end{pmatrix}. Prove by mathematical induction that An=(2nn2n−102n)A^n=\begin{pmatrix}2^n&n2^{n-1}\\0&2^n\end{pmatrix} for every positive integer nn.

(6)

(Total for Question 1 is 6 marks)

Your progress and exam materials

This section: Evidence from your answers: 0/1 secureYour confidence: 0 self-rated secureTracker status: 0/1 secure, 0 shaky, 1 unseen

Overall: Evidence from your answers: 0/158 secureYour confidence: 0 self-rated secureTracker status: 0/158 secure, 0 shaky, 158 unseen

Progress is saved on this device for guests and accounts right now; cross-device account sync is not live yet.

See everything due today — your streak and what to revise next, kept on this device.

Want help turning these notes into marks?

Bring a tricky specification point or a recent answer, and we can work through the method and exam wording together.