Pearson Edexcel IAL Mathematics FP1.8.1 Proof by mathematical induction

Practise mathematical induction proofs across summations, divisibility, recurrence definitions and powers of 2x2 matrices.

Exam points

verify the base case explicitly before assuming the statement for n = k
use the recurrence, added term or matrix multiplication to derive the n = k + 1 case
for divisibility proofs, rewrite f(k + 1) using f(k) plus a clear multiple of the divisor

[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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FP1.8.1 - Proof by mathematical induction question 1