Home → Curriculum Hub → Maths Lessons → Schemes of Work → A-Level Further Mathematics → Proof by Induction
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.
By the end of this unit students will be able to:
Students should be secure with the following before beginning this unit.
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.
Replace the sum up to \(k\) with the formula for \(k\), then rearrange until the formula for \(k + 1\) appears.
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,
Substitute the result for \(A^{k}\) before you multiply. The conclusion names every integer from the base case on.
| Misconception | Teaching 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. |
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