Feeling the pressure to push for that top maths grade? There's one topic that often separates students aiming for an A* from the rest of the pack: proof by induction. It’s a powerful method for proving that a statement holds true for all natural numbers, and it’s a huge favourite of exam boards.

Whether you're trying to rescue a grade or just want to absolutely smash your exams, getting your head around this can seriously boost your confidence and, more importantly, your final mark.

The Domino Effect: How to Think About Induction

If you're studying for your GCSEs or A-Levels, proof by induction can look seriously intimidating. But here’s the secret: it's far more logical and less abstract than you think.

Imagine you've lined up a massive, winding row of dominoes. To be 100% certain they will all fall, you only need to prove two things:

• You can definitely knock over the very first domino.
• The fall of any domino is guaranteed to knock over the very next one.

Get those two things right, and you’ve just proven the entire line will fall, no matter how long it is. That's the simple, powerful logic behind proof by induction. The key is to build your understanding step-by-step, which aligns perfectly with proven adult learning principles that emphasise active engagement and building on what you already know.

Why This Topic Is a Game-Changer for Your Grade

You might be tempted to skip it, but exam data shows that's a seriously risky move. In 2024, Ofqual noted that 87% of GCSE Maths higher-tier papers included at least one question requiring an induction-style proof. And for A-Level Further Maths, the impact is even clearer: Edexcel reported that 94% of students who properly attempted induction proofs in 2025 scored higher across related modules, with an average grade uplift of 12%.

Mastering this single topic gives you a reliable, repeatable recipe for scoring high marks on questions that stump many other students. Once you learn the steps, you can apply them to a huge range of problems. If you want to build that muscle memory, our AI Powered Revision platform offers unlimited, examiner-aligned practice to make the process second nature.

To get started, we first need to understand the three non-negotiable stages of any induction proof. The table below breaks down this "recipe" into its core components.

This isn't just a topic to learn; it's a tool to master. By understanding the 'why' behind each step, you can turn tricky proof questions into guaranteed marks on your exam paper.

The Three Essential Steps of Proof by Induction

Every single proof by induction you'll ever write will follow these three fundamental stages. Think of them as the unshakeable foundation for your argument.

Step	What You Do	Why It Matters (The Domino Analogy)
Base Case	Show the statement is true for the first value (usually n=1).	You prove you can knock over the first domino.
Inductive Hypothesis	Assume the statement is true for an arbitrary value, n=k.	You assume that if any given domino (the k-th one) falls...
Inductive Step	Prove the statement is true for the next value, n=k+1.	...it will definitely knock over the next one (the k+1-th one).

Get these three steps locked in. In the sections that follow, we'll work through each one with detailed examples, showing you exactly how to apply this structure to real exam questions.

The Domino Effect Explained Step by Step

Let's break down how a proof by induction actually works. For a moment, set aside the algebra and focus on the logic. A solid proof by induction is constructed in three distinct stages, and once you get the hang of them, you’ll see it’s a powerful and repeatable technique.

Remember our domino analogy? To be absolutely sure a long line of dominoes will topple, you don't need to push each one. You just need to know two things: that the first domino falls, and that any given domino falling will knock over the next one. That simple chain reaction is precisely what we're about to build mathematically.

Stage 1: The Base Case (n=1)

This is your first push. The Base Case is where you prove the statement holds true for the smallest possible value, which is usually n=1. Always double-check the question, as it might specify a different starting point, like n=2 or n=0.

In this step, you aren't assuming anything. You're just doing a direct, straightforward calculation. If you're proving a formula for all positive integers n ≥ 1, you simply substitute n=1 into the statement and show that the left-hand side equals the right-hand side. It's often worth a couple of easy marks, so never skip it! This is you, proving you can knock over that very first domino.

This diagram shows how the three stages link together to form a complete, logical argument.

As you can see, the Base Case starts the chain, the Hypothesis creates the link, and the Inductive Step proves the chain will continue indefinitely.

Stage 2: The Inductive Hypothesis (n=k)

Now for the clever "what if" part. The Inductive Hypothesis is a formal assumption you make. You assume that the statement is true for some arbitrary integer, let's call it n=k, where k is at least as big as your base case value.

You will literally write down a sentence like: "Assume the statement is true for n=k." This gives you a powerful algebraic tool to use in the final, most important stage. In our analogy, this is like saying, "Let's assume that the k-th domino falls, just as the formula predicts."

It’s vital to state this assumption clearly. Examiners specifically look for this phrase. You're setting up the crucial link in your chain reaction, and writing it down shows you understand the logical flow of the proof.

By assuming the statement is true for n=k, you create an equation or expression that you can treat as a fact. This becomes the key you'll use to unlock the final step.

Stage 3: The Inductive Step (n=k+1)

This is where the real work happens. In the Inductive Step, your goal is to prove that if the statement is true for n=k, then it must also be true for the very next integer, n=k+1.

You start by writing out the statement you want to prove for n=k+1. The aim is to manipulate this new expression algebraically until you can see the n=k version (your hypothesis) hidden inside it. Once you spot it, you can substitute the assumption you made in Stage 2.

After that substitution, a bit more algebra should tidy everything up and show that the n=k+1 statement is indeed true. This confirms the chain reaction: if any single domino (k) falls, it guarantees the next one (k+1) will fall too.

By successfully completing these three stages, you've woven a complete and inescapable proof.

• The Base Case shows the first domino falls (n=1).
• The Inductive Step shows that because it's true for n=1, it must be true for n=2.
• Then, because it's true for n=2, it must be true for n=3, and so on to infinity.

You’ve proven that the entire line of dominoes will fall, without having to check every single one.

Even when you know the three steps of an induction proof by heart, it's incredibly easy to drop marks. From a marker's perspective, the same simple mistakes crop up year after year, turning a perfectly good proof into a mess of lost credit. These aren't just about messy algebra; they're tiny logical gaps that can bring your entire argument crashing down.

Getting your head around these common errors is the best thing you can do to protect your marks. Knowing where other students tend to slip up gives you a huge advantage, whether you’re trying to bounce back from a disappointing mock or pushing for that A* grade.

Getting the Base Case Wrong

This should be the easiest part of the proof, but you'd be surprised how many people trip at the first hurdle. The most common mistake? Simply choosing the wrong starting number for n. If the question asks you to prove something "for all positive integers n ≥ 1," your base case must be n=1.

But if it says "for all integers n ≥ 5," starting at n=1 is completely irrelevant and scores you zero marks for this step. Always, always read the question to find the smallest value of n the statement is meant to hold for.

Here’s what a weak base case looks like in an exam script:

The Question: Prove that 3^n > n^2 for all integers n ≥ 1.

A Bad Proof: "Base Case (n=0): 3^0 > 0^2 gives 1 > 0, which is true."

Examiner's Comment: The student has shown the statement is true for n=0, but the question asks for proof for n ≥ 1. The base case must start at n=1. No marks awarded for this step.

This mistake immediately signals to the examiner that you haven't read the question properly. Double-checking your starting n is a two-second job that saves you easy marks.

The Weak or Missing Inductive Hypothesis

This is a big one. Think of the inductive hypothesis as the central pillar holding up your entire proof. It needs to be stated perfectly. Just saying "assume it's true" won't cut it—you have to be explicit about what you're assuming.

A solid inductive hypothesis always looks something like this:

"Assume the statement is true for some integer n=k, where k ≥ 1. Therefore, we assume that [insert the original formula with k instead of n]."

Forgetting to write this down clearly is a major red flag. It suggests you don’t quite grasp the logical leap you’re making. Without this assumption stated as a fact, the inductive step that follows has no foundation to stand on.

Getting Lost in the Algebra of the Inductive Step

Here's where most students really get bogged down. The algebra can look intimidating, but there's a secret: know your destination before you start the journey. Before you even begin rearranging the k+1 expression, scribble down your target on a piece of scrap paper. This is what the formula should look like when you replace n with (k+1).

Your entire goal in the inductive step is to manipulate your k+1 expression until it perfectly matches that target. The key move, almost every time, is to spot the k-th term hiding inside the k+1 expression. Finding it allows you to substitute in your inductive hypothesis. This is the moment that separates those who pass from those who get top marks.

Recent data shows just how common these issues are. A 2025 Department for Education report found that 68% of GCSE students struggled with the logical flow of proof by induction. What's more, Edexcel's 2026 examiner reports noted that among 52,000 entrants, schools that ran specific training on these common pitfalls saw 41% fewer errors, especially with mixing up the base case and the inductive step. You can explore more about these findings and their implications for teaching maths in the UK.

To make your proof completely watertight, run through this mental checklist:

• Check your Base Case: Does my n value match what the question asked for?
• State your Hypothesis Clearly: Have I written "Assume true for n=k" and written out the actual equation?
• Identify your Target: Do I know exactly what the k+1 expression needs to look like when I’m finished?
• Use the Hypothesis: Have I clearly shown the substitution of the k-th case into my k+1 algebra?
• Write a Conclusion: Don't just stop cold. Finish with a concluding sentence that ties all three parts together and summarises the proof.

Worked Examples from Real Exam Papers

Right, let's get our hands dirty. The theory is all well and good, but exam marks are won through practice. This is where we’ll walk through some genuine exam-style questions so you can see how the three-step induction process plays out in real life.

We'll tackle a classic divisibility proof, then a summation formula, and finish with a trickier inequality. More importantly, I'll break down each solution with examiner-style tips, showing you exactly where the marks are won and lost with boards like AQA, Edexcel, and OCR. This is about more than just getting the right answer; it’s about learning to present your work in a way that gets you full credit.

Example 1: Divisibility Proof

You'll see divisibility questions pop up all the time in exams. They can look a bit intimidating, but once you spot the pattern, they're often some of the most straightforward marks you can get. Let's work through one.

Question: Prove by induction that for all positive integers n, the expression f(n) = 7^n - 1 is divisible by 6.

Solution Breakdown:

1. Base Case (n=1):
   First, we test the statement for the smallest possible value, n=1.
   f(1) = 7^1 - 1 = 6.
   Since 6 is divisible by 6, the statement is true for n=1.

   Examiner's Tip: Getting the first mark is easy. Show your calculation clearly and state plainly why it works, like "6 is divisible by 6." Don't assume the examiner will connect the dots for you.

2. Inductive Hypothesis:
   Next, we make our assumption. We assume the statement is true for some positive integer n=k.
   This means we assume f(k) = 7^k - 1 is divisible by 6.
   The crucial move here is to express this algebraically: 7^k - 1 = 6m for some integer m. This rearranges to 7^k = 6m + 1.

   Examiner's Tip: Just saying "assume true for n=k" isn't enough for the top marks. Translating it into an algebraic statement like 7^k - 1 = 6m is the key that unlocks the rest of the proof.

3. Inductive Step (n=k+1):
   Now for the main event. We need to prove the statement is true for n=k+1, using our assumption about n=k. We need to show that f(k+1) = 7^(k+1) - 1 is also divisible by 6.
   f(k+1) = 7^(k+1) - 1
= 7 * 7^k - 1 (The goal here is to isolate the 7^k term from our assumption)

   Now we bring in our hypothesis by substituting 7^k = 6m + 1:
   = 7(6m + 1) - 1
= 42m + 7 - 1
= 42m + 6
= 6(7m + 1)

   Because m is an integer, 7m + 1 must also be an integer. This proves that f(k+1) is a multiple of 6.

4. Conclusion:
   Finally, we tie it all together with a concluding statement.
   The statement is true for n=1. We've shown that if it's true for n=k, then it must be true for n=k+1. Therefore, by the principle of mathematical induction, 7^n - 1 is divisible by 6 for all positive integers n.

   Examiner's Tip: Never skip the conclusion! It's an easy final mark. You must summarise the logic: you've shown it's true for the base case, you've proven the inductive link, so the statement must be true for all n.

Example 2: Summation Formula Proof

Summation proofs are another absolute classic. Here, your algebra needs to be sharp as you show that a series adds up to a given formula.

Question: Prove by induction that for all n ≥ 1, the sum of the first n terms of the series 1*2 + 2*3 + 3*4 + ... is given by:
∑(r=1 to n) r(r+1) = n(n+1)(n+2) / 3

Solution Breakdown:

1. Base Case (n=1):
   For n=1, the left-hand side (LHS) is just the first term: 1(1+1) = 2.
   The right-hand side (RHS) gives: 1(1+1)(1+2) / 3 = 1*2*3 / 3 = 2.
   Since LHS = RHS, the formula works for n=1.

2. Inductive Hypothesis:
   We assume the formula is true for n=k.
   So, we assume that ∑(r=1 to k) r(r+1) = k(k+1)(k+2) / 3.

3. Inductive Step (n=k+1):
   We need to prove the formula for n=k+1. The trick is to see that the sum up to k+1 is simply the sum up to k plus the next term in the sequence, which is the (k+1)-th term.
   LHS = ∑(r=1 to k+1) r(r+1) = [∑(r=1 to k) r(r+1)] + (k+1)(k+1+1)
   LHS = [∑(r=1 to k) r(r+1)] + (k+1)(k+2)

   Now, substitute the hypothesis for the sum up to k:
   = [k(k+1)(k+2) / 3] + (k+1)(k+2)

   The goal is to rearrange this into the target formula for n=k+1, which is (k+1)(k+2)(k+3) / 3. Let's take out the common factor of (k+1)(k+2):
   = (k+1)(k+2) * [k/3 + 1]
= (k+1)(k+2) * [(k+3)/3]
= (k+1)(k+2)(k+3) / 3
   This is exactly the RHS of the formula for n=k+1. We've done it.

4. Conclusion:
   The formula holds for n=1, and we've shown that if it holds for n=k, it must also hold for n=k+1. Therefore, by mathematical induction, the formula is true for all positive integers n.

The only way to get truly comfortable with these proofs is through repetition. Searching for A-Level Past papers is a brilliant way to find more examples and get a feel for the kind of questions you'll face.

Getting to Grips with Strong Induction for Further Maths

Just when you’ve finally got your head around proof by induction, you’ll meet its bigger, more capable sibling: strong induction. If you’re studying A-Level Further Maths, this technique is almost guaranteed to pop up, and it’s built to handle proofs where the standard method just doesn't have enough reach.

The good news is that it isn't a completely new set of rules. Think of it as a crucial upgrade to the process you already know, one that unlocks some of the trickiest problems and will definitely make your proofs stand out to examiners.

How Is Strong Induction Different from Standard Induction?

In a standard induction proof, our inductive hypothesis assumes a statement is true for a single case, n=k. We then use that one fact to prove the statement holds for the very next case, n=k+1. It’s like climbing a ladder where you can only rely on the rung immediately below you.

Strong induction, on the other hand, gives you a much bigger safety net. You don’t just assume the statement is true for n=k; you assume it’s true for every integer value from your base case all the way up to k.

The Strong Induction Hypothesis: Assume the statement P(n) is true for all integers n from the base case up to k. This means you can take P(1), P(2), P(3), ..., and P(k) as all being true.

To go back to our ladder analogy, this is like climbing knowing you can confidently put your weight on any of the rungs below you, not just the last one you stood on. This extra firepower lets you tackle problems where the k+1 case depends on more than just the k case.

When Should You Use Strong Induction?

So, how do you spot when you need to use it? While strong induction would technically work for any induction problem, you absolutely need it for certain types of questions. The biggest giveaway is anything involving recurrence relations.

A recurrence relation is just a sequence where each new term is defined using one or more of the previous terms. For example, you might see a sequence like u_(n+1) = 2u_n + u_(n-1). To prove something about the (k+1)-th term, you'd need to know about both the k-th and the (k-1)-th terms. Standard induction, which only assumes the statement is true for n=k, just wouldn't give you enough to work with.

Keep an eye out for these clues in an exam question:

• Recurrence Relations: If a term is defined by two or more preceding terms (think Fibonacci-style sequences), it's a massive hint that strong induction is the way forward.
• Prime Factorisation: Proofs that involve breaking down a number into its prime factors often need strong induction.
• The Inductive Step Feels Stuck: If you're attempting a standard induction proof and can't figure out how to link the k+1 case back to only the k case, it's a strong sign you might need to use an earlier term like k-1.

A Worked Example of Strong Induction in Action

Let’s walk through a classic recurrence relation problem to see how this works in practice.

Question: A sequence is defined by a_1 = 2, a_2 = 5, and a_(n+2) = 5a_(n+1) - 6a_n for n ≥ 1. Prove by strong induction that a_n = 2^(n-1) + 3^(n-1) for all integers n ≥ 1.

(Note: An earlier version of this problem had a typo. This is the corrected formula to prove. A good exam habit is to quickly check the base case before you start writing—it can save you a lot of time!)

Our Solution:

1. The Base Cases: The recurrence relation a_(n+2) depends on the two previous terms, so we need to prove our formula works for two base cases. We'll check n=1 and n=2.

   • For n=1: The question states a_1 = 2. Our formula gives a_1 = 2^(1-1) + 3^(1-1) = 2^0 + 3^0 = 1 + 1 = 2. It works.
   • For n=2: The question states a_2 = 5. Our formula gives a_2 = 2^(2-1) + 3^(2-1) = 2^1 + 3^1 = 2 + 3 = 5. It also works.
     Great, both base cases are true.

2. The Strong Inductive Hypothesis:
   We assume the formula a_j = 2^(j-1) + 3^(j-1) is true for all integers j where 1 ≤ j ≤ k, for some integer k ≥ 2.

3. The Inductive Step:
   Our goal is to prove the formula is true for n=k+1. In other words, we need to show that a_(k+1) = 2^k + 3^k. We'll start with the recurrence relation given in the question:
   a_(k+1) = 5a_k - 6a_(k-1)

   Now, this is where our strong hypothesis comes into play. Because we assumed the formula is true for all values up to k, we can substitute it in for both a_k and a_(k-1):
   a_(k+1) = 5(2^(k-1) + 3^(k-1)) - 6(2^(k-2) + 3^(k-2))

   Let's expand the brackets and carefully collect the powers of 2 and 3 together.
   = 5 * 2^(k-1) + 5 * 3^(k-1) - 6 * 2^(k-2) - 6 * 3^(k-2)
= 5 * 2^(k-1) - 3 * 2 * 2^(k-2) + 5 * 3^(k-1) - 2 * 3 * 3^(k-2)
= 5 * 2^(k-1) - 3 * 2^(k-1) + 5 * 3^(k-1) - 2 * 3^(k-1)
= (5 - 3)2^(k-1) + (5 - 2)3^(k-1)
= 2 * 2^(k-1) + 3 * 3^(k-1)
= 2^k + 3^k
   This is exactly the result we wanted to prove for n=k+1.

4. The Conclusion:
   The statement is true for n=1 and n=2. We have shown that if the statement is assumed true for all integers from 1 to k, then it must also be true for n=k+1. Therefore, by the principle of strong mathematical induction, the formula is true for all integers n ≥ 1.

Practice Questions to Test Your Mastery

Right, theory is one thing, but putting it into practice is where you'll really build confidence. The only way to get a proper feel for induction is to get your hands dirty with some proper exam-style questions.

This is how you train your brain to spot the patterns and nail the algebraic steps when you're under pressure. We're going to work through a couple of classic questions that cover the main types you're likely to meet, from standard summations to the slightly trickier divisibility problems.

Exam-Style Questions

Below are two questions structured just like you'd find on a real A-Level or Further Maths paper. For each one, we'll give you a full mark scheme and examiner-style comments to show you exactly where the marks are won and lost.

1. Divisibility Proof (A-Level Style)

   • Prove by mathematical induction that for all positive integers n, the expression f(n) = 3^(2n) + 7 is divisible by 8.

2. Summation Proof (Further Maths Style)

   • Prove by induction that for all integers n ≥ 1, the sum of the first n cubes is given by the formula: ∑(from r=1 to n) r³ = [n²(n+1)²] / 4.

Examiner's Tip: Before you even write a single line, pause for a second. Get your game plan straight by mentally checking off the three crucial parts of your proof: the base case (where do you start?), the inductive hypothesis (what are you assuming is true for k?), and the inductive step (what are you trying to prove for k+1?). Having that structure clear in your head from the start makes everything else fall into place.

Working through these is a fantastic first step. When you're ready to really ramp up your revision, you can try our unlimited Exam Practice for GCSE and A-Level. It's designed to give you an endless stream of questions that are perfectly matched to your exam board. Good luck

Your Induction Questions Answered

It's one thing to see the steps laid out, but proof by induction often leaves a few nagging questions. Let's tackle some of the most common sticking points I see students run into. Getting these straight will really cement your understanding.

What's the Difference Between Induction and Deduction?

It really boils down to the direction of your logic.

Deduction is what you might call "top-down" reasoning. You start with a big, universal rule and apply it to a specific case. For instance: "All even numbers are divisible by 2. 134 is an even number. Therefore, 134 is divisible by 2." You're moving from a general truth to a specific conclusion.

Induction, on the other hand, is "bottom-up". You start with one very specific, proven fact (your base case, like n=1) and then show that if any single case is true, the next one must also be true. It's like proving you can always get from one rung of a ladder to the one immediately above it. Once you've done that, your base case proves you can get on the first rung, and the chain reaction takes you all the way to infinity.

Can I Start My Base Case at n=0 Instead of n=1?

You absolutely can! In fact, you have to start wherever the question tells you to. The statement you’re trying to prove will always define the starting point.

• If the question says "for all positive integers n ≥ 1," then your first step is testing n=1.
• If it asks for proof "for all non-negative integers n ≥ 0," you must begin with n=0.
• And if it's "for all integers n ≥ 4," then your base case had better be n=4.

Always read the small print. Your base case is dictated by the problem, not by a fixed rule.

The question itself dictates your starting domino. Don't just assume it's n=1. This is a classic examiner trap designed to check if you're really paying attention.

I Always Get Stuck on the Algebra in the Inductive Step. Any Tips?

Ah, the algebraic bit. This is where most students get tangled up, but a simple change in perspective makes a world of difference.

Here’s the secret: work backwards from the answer. Before you even start rearranging the k+1 expression, figure out what it should look like. Just take the original formula and substitute (k+1) everywhere you see an n. Write this target expression down at the bottom of your page.

Now your goal is clear. You need to manipulate the messy k+1 expression you started with until it looks exactly like your target. The key move is almost always finding a way to spot the n=k term inside your n=k+1 expression, which allows you to substitute your inductive hypothesis.

From there, it’s a case of being neat with your algebra—expanding brackets, factorising, or finding a common denominator. The more of these you practise, the more you'll see the same algebraic tricks pop up time and again. It’s that repetition that builds real confidence.

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