Contoh Soal 2 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{l}\\ 6.& \text{Dengan Induksi Matematika untuk}n\in \mathbb{N}\\ & \text{dapat dibuktikan juga bahwa}{7}^n-2^n\\ & \text{akan habis dibagi oleh}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}P(n)& =7^n-2^n\\ P(1)& =7^1-2^1\\ & =7-2=5\\ & \text{adalah bilangan yang habis dibagi 5}\end{array}\end{array}$

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

$\begin{array}{l}\\ 8.& \text{Bila kita hendak membuktikan}{\displaystyle \sum _{i=1}^n={\displaystyle \frac{1}{2}n(n+1)}}\\ & \text{dengan induksi matematika}\\ & \text{maka untuk langkah}n=k+1\\ & \text{bentuk yang harus ditunjukkan adalah}...\\ & \begin{array}{l}\\ \text{a}.& 1+2+3+\cdots +n={\displaystyle \frac{1}{2}n(n+1)}\\ \text{b}.& 1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+1)}\\ \text{c}.& 1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+2)}\\ \text{d}.& 1+2+3+\cdots +k+(k+1)={\displaystyle \frac{1}{2}(k+1)(k+2)}\\ \text{e}.& 1+2+3+\cdots +k+(k+1)={\displaystyle \frac{1}{2}(k+2)(k+3)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}P(n)& =1+2+3+\cdots +n={\displaystyle \frac{1}{2}n(n+1)}\\ P(k)& =1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+1)}\\ P(k+1)& =1+2+3+\cdots +k+(k+1)\\ & =\underset{{\displaystyle \frac{1}{2}k(k+1)}}{\underbrace{1+2+3+\cdots +k}}+(k+1)\\ & ={\displaystyle \frac{1}{2}k(k+1)+(k+1)=(k+1)\left({\displaystyle \frac{1}{2}k+1}\right)}\\ & =(k+1){\displaystyle \frac{1}{2}(k+2)}\\ & ={\displaystyle \frac{1}{2}(k+1)(k+2)}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika}P(n)={\displaystyle \frac{n-1}{n+3},\text{maka}P(k+1)}\\ & \text{dinyatakan dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{k-1}{k+3}}\\ \text{b}.& {\displaystyle \frac{k-1}{k+4}}\\ \text{c}.& {\displaystyle \frac{k}{k+4}}\\ \text{d}.& {\displaystyle \frac{k+1}{k+4}}\\ \text{e}.& {\displaystyle \frac{k+1}{k+5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}P(n)=& {\displaystyle \frac{n-1}{n+3}}\\ P(k+1)& ={\displaystyle \frac{k+1-1}{k+1+3}}\\ & ={\displaystyle \frac{k}{k+4}}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika}P(n)={\displaystyle \frac{n^2+1}{4},\text{maka}}\\ & \text{pernyataan untuk}P(k+1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{k^2+2k+1}{4}}\\ \text{b}.& {\displaystyle \frac{k^2+2k+2}{4}}\\ \text{c}.& {\displaystyle \frac{k^2+2k+2}{5}}\\ \text{d}.& {\displaystyle \frac{k^2+2k+3}{5}}\\ \text{e}.& {\displaystyle \frac{k^2+2k+3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}P(n)& ={\displaystyle \frac{n^2+1}{4}}\\ P(k+1)& ={\displaystyle \frac{(k+1{)}^2+1}{4}}\\ & ={\displaystyle \frac{k^2+2k+2}{4}}\end{array}\end{array}$