Proof by Induction

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This scheme of work for Core Pure 1 proof by induction builds on the sum formulae for integers, squares and cubes, and on matrix multiplication. It develops the skill of using the case \(n = k\) inside the case \(n = k + 1\). These ideas are used when a result has to be proved for every positive integer \(n\).

Students prove sum formulae first, then divisibility statements. They finish by proving a result about the powers of a matrix.

What Success Looks Like

By the end of this unit students will be able to:

  • Prove a sum formula by induction.
  • Prove a divisibility statement by induction.
  • Prove a result about powers of a matrix.
  • State the conclusion for every valid \(n\).

Prerequisite Knowledge

Students should be secure with the following before beginning this unit.

  • Use the formulae for the sums of the first \(n\) integers, squares and cubes.
  • Factorise an expression to show a constant factor.
  • Multiply two square matrices.

Key Mathematical Ideas

Four steps, in this order

Check the first value of \(n\). Assume the result for \(n = k\). Prove it for \(n = k + 1\) by using that assumption. If it holds at the start, and moves from \(k\) to \(k + 1\), it holds for every integer from the base case on.

Add the next term

Replace the sum up to \(k\) with the formula for \(k\), then rearrange until the formula for \(k + 1\) appears.

\[\sum_{r=1}^{k+1} u_r = \left(\sum_{r=1}^{k} u_r\right) + u_{k+1}\]

Make the factor visible

Assume \(f(k)\) is a multiple of \(m\). Write \(f(k+1)\) so that \(f(k)\) appears, then factorise \(m\) out of every term. For example,

\[3^{k+1} – 1 = 3\left(3^{k} – 1\right) + 2\]

Multiply by the matrix again

Substitute the result for \(A^{k}\) before you multiply. The conclusion names every integer from the base case on.

\[A^{k+1} = A^{k}A\]

Working Mathematically

Fluency

  • Check the case \(n = 1\) for the sum of the first \(n\) integers.
  • Write the term that turns a sum up to \(k\) into a sum up to \(k + 1\).
  • Write \(A^{k+1}\) as \(A^{k}A\).

Reasoning

  • Explain why the base case alone does not prove the statement.
  • Explain why \(f(k)\) must appear in the working for \(f(k+1)\).
  • Explain why the conclusion has to name the values of \(n\).

Common Misconceptions with Proof by Induction

MisconceptionTeaching focus
Checking the base case and then stopping.The base case starts the chain. The inductive step is what moves from \(k\) to \(k + 1\).
Writing the target for \(k + 1\) and treating it as proved.Start from the case \(k\). The algebra has to reach the statement for \(k + 1\).
Never substituting the statement for \(n = k\).The result for \(n = k\) has to appear in the working for \(n = k + 1\).
Leaving a divisibility proof without a common factor.Write \(f(k+1)\) so that \(f(k)\) is visible, then factorise the divisor out of every term.
Multiplying the matrix power without substituting the case \(k\).\(A^{k+1}\) is \(A^{k}A\). Replace \(A^{k}\) before you multiply.

Differentiation

Additional support

  • Label the four steps before any algebra.
  • For a sum, write the next term on its own line.
  • For divisibility, circle the copy of \(f(k)\) inside \(f(k+1)\).

Additional challenge

  • Prove the sum of the first \(n\) cubes by induction.
  • Prove a divisibility result whose expression has two different powers.
  • Prove a formula for the entries of \(A^{n}\).

Teach This Unit with Mr Mathematics

Every lesson in this unit is already built. A Mr Mathematics membership gives you the presentation, worksheet, question generator and video tutorial for all three lessons, alongside the full Key Stage 3, GCSE, IGCSE and A-Level libraries.

A school membership covers every teacher in your department, so the whole scheme of work is resourced from one subscription.

Proof by Induction Lessons


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