Contoh Soal 2 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{ll}\\ 6.&\textrm{Dengan Induksi Matematika untuk}\: \: n\in \mathbb{N}\\ &\textrm{dapat dibuktikan juga bahwa}\: \: 7^{n}-2^{n}\\ &\textrm{akan habis dibagi oleh}\\ &\begin{array}{lllllll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&5\\ \textrm{e}.&6\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}P(n)&=7^{n}-2^{n}\\ P(1)&=7^{1}-2^{1}\\ &=7-2=5\\ &\textrm{adalah bilangan yang habis dibagi 5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui bahwa}\: \: P(n)\: \: \textrm{rumus dari}\\ & 3+6+9+\cdots +3n=\displaystyle \frac{3}{2}n(n+1)\\ &\textrm{maka langkah pertama dengan induksi matematika}\\ &\textrm{dalam pembuktian rumus tersebut adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.&P(n)\: \: \textrm{benar untuk}\: \: n=-1\\ \color{red}\textrm{b}.&P(n)\: \: \textrm{benar untuk}\: \: n=1\\ \textrm{c}.&P(n)\: \: \textrm{benar untuk}\: \: n\: \: \textrm{bilangan bulat}\\ \textrm{d}.&P(n)\: \: \textrm{benar untuk}\: \: n\: \: \textrm{bilangan rasional}\\ \textrm{d}.&P(n)\: \: \textrm{benar untuk}\: \: n\: \: \textrm{bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Langkah}&\: \textrm{awal yang harus ditunjukkan adalah}\\ n=1&\: \: \textrm{atau}\: \: P(1)\: \: \textrm{harus benar, yaitu}:\\ P(1)&=3.1(\textit{ruas kiri})=\displaystyle \frac{3}{2}.1.(1+1)(\textit{ruas kanan})=3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Bila kita hendak membuktikan}\: \: \displaystyle \sum_{i=1}^{n}=\displaystyle \frac{1}{2}n(n+1)\\ &\textrm{dengan induksi matematika}\\ &\textrm{maka untuk langkah}\: \: n=k+1\\ &\textrm{bentuk yang harus ditunjukkan adalah}...\\ &\begin{array}{lll}\\ \textrm{a}.&1+2+3+\cdots +n=\displaystyle \frac{1}{2}n(n+1)\\ \textrm{b}.&1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+1)\\ \textrm{c}.&1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+2)\\ \color{red}\textrm{d}.&1+2+3+\cdots +k+(k+1)=\displaystyle \frac{1}{2}(k+1)(k+2)\\ \textrm{e}.&1+2+3+\cdots +k+(k+1)=\displaystyle \frac{1}{2}(k+2)(k+3)\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}P(n)&=1+2+3+\cdots +n=\displaystyle \frac{1}{2}n(n+1)\\ P(k)&=1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+1)\\ P(k+1)&=1+2+3+\cdots +k+(k+1)\\ &=\underset{\displaystyle \frac{1}{2}k(k+1)}{\underbrace{1+2+3+\cdots +k}}+(k+1)\\ &=\displaystyle \frac{1}{2}k(k+1)+(k+1)=(k+1)\left ( \displaystyle \frac{1}{2}k+1 \right )\\ &=(k+1)\displaystyle \frac{1}{2}(k+2)\\ &=\displaystyle \frac{1}{2}(k+1)(k+2) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: P(n)=\displaystyle \frac{n-1}{n+3},\: \textrm{maka}\: \: P(k+1)\\ & \textrm{dinyatakan dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{k-1}{k+3}\\ \textrm{b}.&\displaystyle \frac{k-1}{k+4}\\ \color{red}\textrm{c}.&\displaystyle \frac{k}{k+4}\\ \textrm{d}.&\displaystyle \frac{k+1}{k+4}\\ \textrm{e}.&\displaystyle \frac{k+1}{k+5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}P(n)=&\displaystyle \frac{n-1}{n+3}\\ P(k+1)&=\displaystyle \frac{k+1-1}{k+1+3}\\ &=\displaystyle \frac{k}{k+4} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: P(n)=\displaystyle \frac{n^{2}+1}{4},\: \: \textrm{maka}\\ &\textrm{pernyataan untuk}\: \: P(k+1)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{k^{2}+2k+1}{4}\\ \color{red}\textrm{b}.&\displaystyle \frac{k^{2}+2k+2}{4}\\ \textrm{c}.&\displaystyle \frac{k^{2}+2k+2}{5}\\ \textrm{d}.&\displaystyle \frac{k^{2}+2k+3}{5}\\ \textrm{e}.&\displaystyle \frac{k^{2}+2k+3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}P(n)&=\displaystyle \frac{n^{2}+1}{4}\\ P(k+1)&=\displaystyle \frac{(k+1)^{2}+1}{4}\\ &=\displaystyle \frac{k^{2}+2k+2}{4} \end{aligned} \end{array}$