2023TrialShort answer

Q11 (b) (ii)

3 marks

Question

A polygonal number is an integer which can be represented as a series of dots arranged
in the shape of a regular polygon. Triangular numbers, square numbers and pentagonal
numbers are examples of polygonal numbers.

For an $r$ -sided regular polygon, where $r\in\mathbb{Z}^+$ , $r\geq 3$ , the $n$ th polygonal number $P_r(n)$
is given by

P_r(n)=\frac{(r-2)n^2-(r-4)n}{2}

where $n\in\mathbb{Z}^+$ . Hence, the $n$ th triangular number can be expressed as

P_3(n)=\frac{n^2+n}{2}.

The $n$ th polygonal number, $P_r(n)$ , can be represented by the series

P_r(n)=\sum_{m=1}^{n}\left(1+(m-1)(r-2)\right),

where $r\in\mathbb{Z}^+$ , $r\ge 3$ .

Use mathematical induction to prove that

P_r(n)=\frac{(r-2)n^2-(r-4)n}{2}

where $n\in\mathbb{Z}^+$ .

Try the question before revealing the answer.

Solution

ii. Base case: $n=1$

LHS=P_r(1)=\sum_{m=1}^{1}(1+(m-1)(r-2))=1+0(r-2)=1

RHS=\frac{(r-2)\times1^2-(r-4)\times1}{2}=\frac{2}{2}=1=\text{LHS}

Therefore the base case is true.

Assume for some $b\ge1$ , $b\in\mathbb Z$ , that

P_r(b)=\frac{(r-2)b^2-(r-4)b}{2}\tag{*}

Now we wish to prove that

P_r(b+1)=\frac{(r-2)(b+1)^2-(r-4)(b+1)}{2}

\begin{aligned} LHS&=P_r(b+1)\\ &=\sum_{m=1}^{b+1}(1+(m-1)(r-2))\\ &=\sum_{m=1}^{b}(1+(m-1)(r-2))+[1+b(r-2)]\\ &=P_r(b)+1+b(r-2)\\ &=\frac{(r-2)b^2-(r-4)b}{2}+1+b(r-2)\\ &\qquad\text{by }(*)\\ &=\frac{(r-2)b^2-(r-4)b+2+2b(r-2)}{2}\\ &=\frac{(r-2)b^2+2b(r-2)+(r-2)- (r-2)+2-(r-4)b}{2}\\ &=\frac{(r-2)(b+1)^2-(r-4)(b+1)}{2}\\ &=RHS \end{aligned}

Therefore the statement is true for $n=b+1$ if it is true for $n=b$ . The statement is true by mathematical induction for integers $n\ge1$ .

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Proof

Further Proof by Mathematical InductionUse mathematical induction for series, divisibility, inequalities, algebra, calculus, probability, geometry and recursive formulae.

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