Contoh Soal 4 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{l}\\ 16.& \text{Diketahui}1+5+9+...+(4n-1)=2n^2-n\\ & \text{dengan}n\text{bilangan asli}.\text{Jika}m<k\text{dengan}\\ & m,k\text{bilangan asli juga},\text{maka}\\ & (4m-3)+(4m+1)+...+(4k-3)=....\\ & \begin{array}{l}\\ \text{a}.& (k-m)(2k+2m-2)\\ \text{b}.& (k-m+1)(2k+2m-3)\\ \text{c}.& (k-m+1)(2k-2m+1)\\ \text{d}.& (k-m+1)(2k^2+2m^2-3)\\ \text{e}.& (k-m{)}^2(2k-2m+4)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& 1+5+9+...+(4m-3)+(4m+1)+...+(4k-3)\\ & =\underset{2k^2-k}{\underbrace{1+5+...+(4k-3)}}-\underset{2(m-1{)}^2-(m-1)}{\underbrace{1+5+...+(4(m-1)-3)}}\\ & =2k^2-k-(2(m-1{)}^2-(m-1))\\ & =2k^2-k-2(m-1{)}^2+(m-1)\\ & =2k^2-k-2(m^2-2m+1)+m-1\\ & =2k^2-k-2m^2+4m-2+m-1\\ & =2k^2-k-2m^2+5m-3\\ & =(k-m+1)(2k+2m-3)\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Diketahui}{2}^1+2^2+2^3+...+2^n=2^{n+1}-2\\ & \text{dengan}n\text{bilangan asli}.\text{Jika}k\text{bilangan asli},\\ & \text{maka}{2}^2+2^3+2^4+...+2^k+2^{k+1}=....\\ & \begin{array}{l}\\ \text{a}.& (k-m)(2k+2m-2)\\ \text{b}.& (k-m+1)(2k+2m-3)\\ \text{c}.& (k-m+1)(2k-2m+1)\\ \text{d}.& (k-m+1)(2k^2+2m^2-3)\\ \text{e}.& (k-m{)}^2(2k-2m+4)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 2^2+2^3+2^4+...+2^k+2^{k+1}\\ & =2^1+2^2+2^3+2^4+...+2^k+2^{k+1}-2^1\\ & =\underset{2^{k+1+1}-2}{\underbrace{2^1+2^2+2^3+2^4+...+2^k+2^{k+1}}}-2^1\\ & =2^{k+2}-2-2\\ & =2^{k+2}-4\\ & =2^k{.2}^2-4\\ & =2^k\times 4-4\\ & =4(2^k-1)\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 2+5+10+17+...+(n^2+1)={\displaystyle \frac{1}{6}(n+1)(2n^2+n+6)}\\ & \text{Jika}S(n)\text{benar, untuk}n=k,\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 2+5+10+17+...+(k^2+1)\\ & ={\displaystyle \frac{1}{6}(k+1)(2k^2+k+6)}\\ \text{b}.& 2+5+10+17+...+(n^2+1)\\ & ={\displaystyle \frac{1}{6}(k+1)(2k^2+k+6)}\\ \text{c}.& 2+5+10+17+...+(k^2+2)\\ & ={\displaystyle \frac{1}{6}(k+2)(2k^2+5k+9)}\\ \text{d}.& (k^2+1)={\displaystyle \frac{1}{6}(k+1)(2k^2+k+6)}\\ \text{e}.& (n^2+2)={\displaystyle \frac{1}{6}(n+1)(2n^2+5n+9)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Cukup jelas}\\ & \text{Tinggal mensubstitusikan dari}\\ & \text{tiap}n\text{diganti}k\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 12+17+22+...+(5n+7)={\displaystyle \frac{1}{2}(n+1)(5n+14)}\\ & \text{Jika}S(n)\text{benar, untuk}n=k,\text{maka benar}\\ & \text{untuk}n=k+1.\text{Pernyataan ini dapat}\\ & \text{dinyatakan dengan}....\\ & \begin{array}{l}\\ \text{a}.& 12+17+22+...+(5k+7)={\displaystyle \frac{1}{2}(k+1)(5k+14)}\\ \text{b}.& 12+17+22+...+(5k+7)={\displaystyle \frac{1}{2}(k+1)(5k+19)}\\ \text{c}.& 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+1)(5k+19)}\\ \text{d}.& 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+2)(5k+14)}\\ \text{e}.& 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+2)(5k+19)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 12+17+22+...+(5k+7)={\displaystyle \frac{1}{2}(k+1)(5k+14)}\\ & 12+17+22+...+(5(k+1)+7)\\ & ={\displaystyle \frac{1}{2}((k+1)+1)(5(k+1)+14)}\\ & 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+2)(5k+19)}\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 4+5+6+7+...+(n+3)<5n^2\\ & \text{Jika}S(n)\text{benar, untuk}n=k+1,\text{maka}\\ & \text{pernyataan ini dapat ditulis dengan}....\\ & \begin{array}{l}\\ \text{a}.& 4+5+6+...+(k+4)<5k^2\\ \text{b}.& 4+5+6+...+(k+3)<5k^2\\ \text{c}.& 4+5+6+...+(k+3)<5(k+1{)}^2\\ \text{d}.& 4+5+6+...+(k+4)<5(k^2+2k+1)\\ \text{e}.& 4+5+6+...+(k+4)<5(k+1)(k-1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 4+5+6+...+(n+3)<5n^2\\ & \text{Saat}n=k+1,\text{maka}\\ & 4+5+6+...+((k+1)+3)<5(k+1{)}^2\\ & =4+5+6+...+(k+4)<5(k^2+2k+1)\end{array}\end{array}$