NSW Y12 Maths - Extension 1 Proof Series

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$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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