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Proof by mathematical inductionEdexcel International A Level Further Maths: Revision notes

Section 1

The four steps of an induction proof

To prove a statement P(n)P(n) for all positive integers nn:

  1. Basis: show P(1)P(1) is true (check both sides of the statement).
  2. Assumption: assume P(k)P(k) is true for some positive integer kk.
  3. Inductive step: use the assumption to show P(k+1)P(k+1) is true.
  4. Conclusion: state that P(1)P(1) is true and P(k)⇒P(k+1)P(k)\Rightarrow P(k+1), so P(n)P(n) is true for all positive integers nn by mathematical induction. The idea is a chain: P(1)P(1) gives P(2)P(2), which gives P(3)P(3), and so on. The basis starts the chain; the step keeps it going. Without either, the proof is incomplete.
Key termsbasisinductive stepassumptionconclusion
Common mistake

Assuming the result is true for n=k+1n=k+1. You assume n=kn=k and must prove n=k+1n=k+1.

Section 2

Summation of series

The step adds the next term to the assumed sum. Example: prove ∑r=1nr(r+1)=n(n+1)(n+2)3\sum_{r=1}^{n}r(r+1)=\frac{n(n+1)(n+2)}{3}. For n=1n=1: LHS =2=2 and RHS =1×2×33=2=\frac{1\times2\times3}{3}=2. Assume true for n=kn=k. Then ∑r=1k+1r(r+1)=k(k+1)(k+2)3+(k+1)(k+2)=(k+1)(k+2)(k+3)3,\sum_{r=1}^{k+1}r(r+1)=\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=\frac{(k+1)(k+2)(k+3)}{3}, which is the formula with n=k+1n=k+1. The same method proves ∑r=1nr3=n2(n+1)24\sum_{r=1}^{n}r^3=\frac{n^2(n+1)^2}{4}. In the step, factorise out common factors rather than expanding everything.

Key termssummationcommon factor
Exam tip

Write down the target, P(k+1)P(k+1), before you start so you know the form you are heading for.

Section 3

Divisibility

To prove f(n)f(n) is divisible by mm: show f(1)f(1) is divisible by mm, assume f(k)=m×integerf(k)=m\times\text{integer}, and then express f(k+1)f(k+1) using f(k)f(k). Example: f(n)=32n+1+1f(n)=3^{2n+1}+1 is divisible by 44. f(1)=28=4×7f(1)=28=4\times7. Then f(k+1)−9f(k)=32k+3+1−9(32k+1+1)=−8f(k+1)-9f(k)=3^{2k+3}+1-9\left(3^{2k+1}+1\right)=-8, so f(k+1)=9f(k)−8f(k+1)=9f(k)-8. Both terms are multiples of 44, so f(k+1)f(k+1) is divisible by 44. Another route: if f(k)=4mf(k)=4m then 32k+1=4m−13^{2k+1}=4m-1 and f(k+1)=9(4m−1)+1=4(9m−2)f(k+1)=9(4m-1)+1=4(9m-2).

Key termsdivisiblemultiple
Exam tip

Look for f(k+1)−a f(k)f(k+1)-a\,f(k) where aa is chosen to cancel the power; the remainder should be a constant multiple of mm.

Section 4

General terms of a sequence

If a sequence is defined by un+1=g(un)u_{n+1}=g(u_n) with u1u_1 given, a formula for unu_n is proved by induction. Example: un+1=3un+4u_{n+1}=3u_n+4 with u1=1u_1=1, prove un=3n−2u_n=3^n-2. For n=1n=1: 3−2=13-2=1. Assume uk=3k−2u_k=3^k-2. Then uk+1=3(3k−2)+4=3k+1−2u_{k+1}=3\left(3^k-2\right)+4=3^{k+1}-2, which is the formula with n=k+1n=k+1. The step always substitutes the assumed formula for uku_k in the recurrence.

Key termsrecurrence relationgeneral term
Common mistake

Starting the induction at the wrong value: use the given first term, and check the formula against it.

Section 5

Matrix powers

For a 2×22\times2 matrix A\mathbf{A}, a formula for An\mathbf{A}^n is proved by writing Ak+1=AkA\mathbf{A}^{k+1}=\mathbf{A}^k\mathbf{A} and multiplying. Example: A=(1101)\mathbf{A}=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} and An=(1n01)\mathbf{A}^n=\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}. For n=1n=1 this is A\mathbf{A}. Assume it for n=kn=k; then Ak+1=(1k01)(1101)=(1k+101)\mathbf{A}^{k+1}=\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}, the formula with n=k+1n=k+1. Always show all four entries, and compare with the formula with nn replaced by k+1k+1.

Key termsmatrix poweridentity
Exam tip

State the conclusion in full every time; examiners award a mark for the complete sentence.

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Carry on to the next subtopic.

Exam questions on Proof by mathematical induction

  1. A student is proving by induction that ∑r=1nr=n(n+1)2\sum_{r=1}^{n}r=\frac{n(n+1)}{2} for all positive integers nn, and has already checked the case n=1n=1.
    Assuming the result is true for n=kn=k, show that it is true for n=k+1n=k+1.2 marks
  2. Let f(n)=7n−1f(n)=7^n-1 for positive integers nn.
    Hence, or otherwise, show that if f(k)f(k) is divisible by 66 then f(k+1)f(k+1) is divisible by 66.2 marks
  3. A sequence is defined by u1=2u_1=2 and un+1=2un+3u_{n+1}=2u_n+3 for n≥1n\ge1.
    Find u2u_2, u3u_3 and u4u_4, and verify that the formula un=5×2n−1−3u_n=5\times2^{n-1}-3 gives the correct values for n=1,2,3,4n=1,2,3,4.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).