Edexcel A-level Further Maths revision notes

Proof

Section CP-1

Year 1

1 specification point

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

Checked against Edexcel 9FM0 section CP-1

Checked against Edexcel 9FM0 section CP-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.

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CP-1.1

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

Notes

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Explanation

Mathematical induction proves a statement $P(n)$ for every integer from a stated starting value. First verify the base case.
Then assume $P(k)$ is true for an arbitrary permitted integer $k$ ; this is the induction hypothesis. Use that hypothesis to derive $P(k+1)$ , rather than assuming the next case.
For a series, add the term with index $k+1$ ; for divisibility, expose the required integer factor; for a matrix power, multiply the assumed expression for $A^k$ by $A$ .
The conclusion must state that the base case and inductive implication together prove $P(n)$ 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 $5^{2n}-1$ is divisible by $24$ for every positive integer $n$ .

1.For $n=1$ , $5^2-1=24$ , so the base case is divisible by $24$ .
2.Assume $5^{2k}-1=24m$ for some positive integer $k$ and some integer $m$ .
3. $5^{2(k+1)}-1=25\left(5^{2k}-1\right)+24$ .
4.Substitution gives $5^{2(k+1)}-1=24(25m+1)$ , which is divisible by $24$ because $25m+1$ is an integer.
5.Therefore $5^{2n}-1$ is divisible by $24$ for every positive integer $n$ .

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

Common mistakes

Don't fall into the trap of assuming the statement for $k+1$ , which makes the inductive step circular.
Don't fall into the trap of writing only that the result is a multiple of $24$ 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 $k+1$ case, and finish with a quantified conclusion.

Tier 1 · Easy

ORIGINAL

Prove by mathematical induction that

1+3+5+\cdots+(2n-1)=n^2

for every positive integer

n

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

Prove by mathematical induction that

8^n-1

is divisible by

7

for every positive integer

n

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

Let

A=\begin{pmatrix}2&1\\0&2\end{pmatrix}

. Prove by mathematical induction that

A^n=\begin{pmatrix}2^n&n2^{n-1}\\0&2^n\end{pmatrix}

for every positive integer

n

(6)

(Total for Question 1 is 6 marks)

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