Pearson Edexcel IAL Mathematics FP1.8 Proof Question Bank

Practise induction proofs for series, divisibility, matrices and sequences, with clear base cases and algebraic k to k + 1 steps.

Exam points

set up the base case and induction hypothesis for sums, matrices or recurrence formulae
transform the k + 1 case to use the assumed result and complete a divisibility or identity proof

Covered syllabus points

FP1.8.1 - Proof by mathematical induction

[Maximum number: 5]

Prove by induction that, for $n \in \mathbb{N}$

\sum_{r=1}^{n} r^{3}=\frac{1}{4} n^{2}(n+1)^{2}

For $n=1, \quad \sum_{r=1}^{1} r^{3}=1$ and $\quad \frac{1}{4}\left(1^{2}\right)(1+1)^{2}=\frac{1}{4} \times 1 \times 4=1$

So true for n=1

B1

(Assume the result is true for n=k, so $\sum_{r=1}^{k} r^{3}=\frac{1}{4} k^{2}(k+1)^{2}$ )
$\text { Then } \quad \sum_{r=1}^{k+1} r^{3}=\frac{1}{4} k^{2}(k+1)^{2}+(k+1)^{3}$

M1

$=\frac{1}{4}(k+1)^{2}\left[k^{2}+4(k+1)\right]=\frac{1}{4}(k+1)^{2}\left[k^{2}+4 k+4\right]$

M1

$=\frac{1}{4}(k+1)^{2}(k+2)^{2}$

A1

Hence result is true for n=k+1. As true for n=1 and have shown if true
for n=k then it is true for n=k+1, so it is true for all $n \in \mathbb{N}$ by induction.

A1

(5)

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Covered syllabus points

FP1.8 - Proof question 1