NSW Y12 Maths - Extension 1 Proof Divisibility

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Divisibility Theory

$The statement $S(n)=m J$ where $J$ is an integer, implies that $S(n)$ is divisible by $m$.\\ The standard rules of mathematical induction are applied as in the example.\\ \begin{multicols}{2} \textbf{Example}\\ %118310 Prove that $3^{3n}+2^{n+2}$ is divisible by $5$ for all positive integers $n$. \\ \textbf{Solution}\\ $\text { To prove } 3^{3 n}+2^{n+2}=5 J \quad(I)$\\ $\begin{aligned} \text{For } n=1 \quad\text { LHS } &=3^3+2^3 \\ &=27+8 \\ &=35 \\ &=5 \mathrm{~J} \quad(J=7)\\ \therefore &(I) \text{ is true for }n=1 \end{aligned}$\\ $\text { Assume (I) is true for } n=k$\\ $\text { i.e. } \left.\quad 3^{3 k}+2^{k+2}=5 J^{\prime} \text { ( } J^{\prime} \text { an integer }\right)\quad \text { (II) }$\\ $\text { Prove (I) is true for } n=k+1$\\ $\text { i.e } 3^{3 k+3}+2^{k+3}=5 J^{\prime \prime}\left(J^{\prime \prime}\right. \text { an integer) } \quad \text{(III)}$\\ \columnbreak $\begin{aligned} \text{In III} \quad L H S&=3^{3 k+3}+2^{k+3}\\ \text{In (II)} \quad 3^{3k}&=5 J^{\prime}-2^2 \times 2^k\\ \text { LHS }&=3^3 \times 3^{3 k}+2^3 \times 2^k\\ &=27\left(5 J^{\prime}-4 \times 2^k\right)+8 \times 2^k\\ &=27 \times 5 J^{\prime}-108 \times 2^k+8 \times 2^k\\ &=27 \times 5J^{\prime}-100 \times 2^k\\ &=5\left(27 J^{\prime}-20 \times 2^k\right)\\ &=5 J^{\prime \prime} \left(J^{\prime \prime}=27 J^{\prime}-20 \times 2^k \text { an integer }\right)\\ &=R H S\\ \therefore LHS&=\text { RHS }\\ &\therefore \text { frue for } n=k+1 \end{aligned}$\\ Now $(I)$ is true for $n=1$ so by mathematical induction (I) is true for all $n \geqslant 1$\\ \end{multicols}$

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