NSW Y12 Maths - Extension 1 Proof Series

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Series Theory

Principle of Mathematical Induction\\  The statement \(S(n)\) may he proved for all positive integers \(n\) by showing that:\\  \textbf{(i) }\(S(1)\) is true; i.e. the statement is true for \(n=1\) (or some other starting point)\\  \textbf{(ii)} If \(S(k)\) is true, then \(S(k+1)\) is true, where \(k \leq n\); i.e. assuming the true for \(n=k\), then it is true for \(n=k+1\). Thus, since \(S(n)\) is true for \(n=1\), it is true for all \(n \geqslant 1\).\\  \begin{multicols}{2}  \textbf{Example}\\ Prove that \(1 \times 2 + 2 \times 3 + \ldots + n(n + 1) = \dfrac{n}{3}(n + 1)(n + 2)\) for all \(n \ge 1\)\\  \textbf{Solution}\\ \(\text { To prove } 1 \times 2+2 \times 3+\cdots+n(n+1)=\dfrac{n}{3}(n+1)(n+2)\quad(I)\)\\  \(\begin{aligned} \text{For }n=1 \qquad \text { LHS } &=1 \times 2 \\ &=2 \\ \text { RHS } &=\dfrac{1}{3}(1+1)(1+2) \\ &=2 \\ \therefore \text { RHS } &=\text { LHS } \\ \therefore (I)& \text { is true for } n=1 \end{aligned}\)\\  \(\text { Assume (I) is true for } n=k\)\\  \(\text {i.e. } 1 \times 2+2 \times 3+\ldots+k(k+1)=\dfrac{k}{3}(k+1)(k+2) \quad (II)\)\\    \columnbreak \(\text { Now prove (I) is true for } n=k+1\)\\  \(\text {i.e. } 1 \times 2+2 \times 3+\ldots+k(k+1)+(k+1)(k+2)=\dfrac{(k+1)}{3}(k+2)(k+3) \quad \text{(III)}\)\\  \(\begin{aligned} \text {In (III) }& \text { LHS }\\ &=1 \times 2+2 \times 3+ \ldots k(k+1)+(k+1)(k+2)\\ &=\dfrac{k}{3}(k+1)(k+2)+(k+1)(k+2) \text { (from II) }\\ &=(k+1)(k+2)\left(\dfrac{k}{3}+1\right)\\ &=(k+1)(k+2) \dfrac{(k+3)}{3} \text {. }\\ &=\dfrac{(k+1)}{3}(k+2)(k+3)\\ &\text { = RHS of (III) }\\ &\therefore \text { LHS =RHS }\\ &\therefore \text { true for } n=k+1 \end{aligned}\)\\  Now (I) is true for \(n=1\), so by mathematical induction (I) is true for all integral \(n \geqslant 1\)\\  \end{multicols}

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  • Series - Video - Lesson 1

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  • Series - Video - Why we need proof by mathematical induction

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  • Series - Video - Proof by Mathematical Induction: Mathematical Dominoes

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  • Series - Video - Proof by induction partial sums worked examples

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