Contoh Soal 4 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{ll}\\ 16.&\textrm{Diketahui}\: \: 1+5+9+...+(4n-1)=2n^{2}-n\\ &\textrm{dengan}\: \: n\: \: \textrm{bilangan asli}.\: \textrm{Jika}\: \: m<k\: \: \textrm{dengan}\\ &m,k\: \: \textrm{bilangan asli juga},\: \textrm{maka}\\ &(4m-3)+(4m+1)+...+(4k-3)=....\\ &\begin{array}{llll}\\ \textrm{a}.&(k-m)(2k+2m-2)\\ \color{red}\textrm{b}.&(k-m+1)(2k+2m-3)\\ \textrm{c}.&(k-m+1)(2k-2m+1)\\ \textrm{d}.&(k-m+1)(2k^{2}+2m^{2}-3)\\ \textrm{e}.&(k-m)^{2}(2k-2m+4) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&1+5+9+...+(4m-3)+(4m+1)+...+(4k-3)\\ &=\underset{2k^{2}-k}{\underbrace{1+5+...+(4k-3)}}-\underset{2(m-1)^{2}-(m-1)}{\underbrace{1+5+...+(4(m-1)-3)}}\\ &=2k^{2}-k-\left ( 2(m-1)^{2}-(m-1) \right )\\ &=2k^{2}-k-2(m-1)^{2}+(m-1)\\ &=2k^{2}-k-2\left ( m^{2}-2m+1 \right )+m-1\\ &=2k^{2}-k-2m^{2}+4m-2+m-1\\ &=2k^{2}-k-2m^{2}+5m-3\\ &=(k-m+1)(2k+2m-3) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Diketahui}\: \: 2^{1}+2^{2}+2^{3}+...+2^{n}=2^{n+1}-2\\ &\textrm{dengan}\: \: n\: \: \textrm{bilangan asli}.\: \textrm{Jika}\: \: k\: \: \textrm{bilangan asli},\\ &\textrm{maka}\: \: 2^{2}+2^{3}+2^{4}+...+2^{k}+2^{k+1}=....\\ &\begin{array}{llll}\\ \textrm{a}.&(k-m)(2k+2m-2)\\ \textrm{b}.&(k-m+1)(2k+2m-3)\\ \textrm{c}.&(k-m+1)(2k-2m+1)\\ \color{red}\textrm{d}.&(k-m+1)(2k^{2}+2m^{2}-3)\\ \textrm{e}.&(k-m)^{2}(2k-2m+4) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&2^{2}+2^{3}+2^{4}+...+2^{k}+2^{k+1}\\ &=2^{1}+2^{2}+2^{3}+2^{4}+...+2^{k}+2^{k+1}-2^{1}\\ &=\underset{2^{k+1+1}-2}{\underbrace{2^{1}+2^{2}+2^{3}+2^{4}+...+2^{k}+2^{k+1}}}-2^{1}\\ &=2^{k+2}-2-2\\ &=2^{k+2}-4\\ &=2^{k}.2^{2}-4\\ &=2^{k}\times 4-4\\ &=4\left ( 2^{k}-1 \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Diketahui bahwa}\: \: S(n)\: \: \textrm{adalah formula dari}\\ &2+5+10+17+...+\left ( n^{2}+1 \right )=\displaystyle \frac{1}{6}(n+1)\left ( 2n^{2}+n+6 \right )\\ &\textrm{Jika}\: \: S(n)\: \: \textrm{benar, untuk}\: \: n=k,\: \: \textrm{maka}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2+5+10+17+...+\left ( k^{2}+1 \right )\\ &=\displaystyle \frac{1}{6}(k+1)\left ( 2k^{2}+k+6 \right )\\ \textrm{b}.&2+5+10+17+...+\left ( n^{2}+1 \right )\\ &=\displaystyle \frac{1}{6}(k+1)\left ( 2k^{2}+k+6 \right )\\ \textrm{c}.&2+5+10+17+...+\left ( k^{2}+2 \right )\\ &=\displaystyle \frac{1}{6}(k+2)\left ( 2k^{2}+5k+9 \right )\\ \textrm{d}.&\left ( k^{2}+1 \right )=\displaystyle \frac{1}{6}(k+1)\left ( 2k^{2}+k+6 \right )\\ \textrm{e}.&\left ( n^{2}+2 \right )=\displaystyle \frac{1}{6}(n+1)\left ( 2n^{2}+5n+9 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Cukup jelas}\\ &\textrm{Tinggal mensubstitusikan dari}\\ &\textrm{tiap}\: \: n\: \: \textrm{diganti}\: \: k \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Diketahui bahwa}\: \: S(n)\: \: \textrm{adalah formula dari}\\ &12+17+22+...+\left ( 5n+7 \right )=\displaystyle \frac{1}{2}(n+1)(5n+14)\\ &\textrm{Jika}\: \: S(n)\: \: \textrm{benar, untuk}\: \: n=k,\: \: \textrm{maka benar}\\ &\textrm{untuk}\: \: n=k+1.\: \textrm{Pernyataan ini dapat}\\ &\textrm{dinyatakan dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&12+17+22+...+\left ( 5k+7 \right )=\displaystyle \frac{1}{2}(k+1)(5k+14)\\ \textrm{b}.&12+17+22+...+\left ( 5k+7 \right )=\displaystyle \frac{1}{2}(k+1)(5k+19)\\ \textrm{c}.&12+17+22+...+\left ( 5k+12 \right )=\displaystyle \frac{1}{2}(k+1)(5k+19)\\ \textrm{d}.&12+17+22+...+\left ( 5k+12 \right )=\displaystyle \frac{1}{2}(k+2)(5k+14)\\ \color{red}\textrm{e}.&12+17+22+...+\left ( 5k+12 \right )=\displaystyle \frac{1}{2}(k+2)(5k+19) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&12+17+22+...+\left ( 5k+7 \right )=\displaystyle \frac{1}{2}(k+1)(5k+14)\\ &12+17+22+...+\left ( 5(k+1)+7 \right )\\ &\qquad\qquad\qquad\quad=\displaystyle \color{magenta}\frac{1}{2}((k+1)+1)(5(k+1)+14)\\ &12+17+22+...+\left ( 5k+12 \right )=\displaystyle \frac{1}{2}(k+2)(5k+19) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Diketahui bahwa}\: \: S(n)\: \: \textrm{adalah formula dari}\\ &4+5+6+7+...+(n+3)<5n^{2}\\ &\textrm{Jika}\: \: S(n)\: \: \textrm{benar, untuk}\: \: n=k+1,\: \: \textrm{maka}\\ &\textrm{pernyataan ini dapat ditulis dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&4+5+6+...+(k+4)<5k^{2}\\ \textrm{b}.&4+5+6+...+(k+3)<5k^{2}\\ \textrm{c}.&4+5+6+...+(k+3)<5(k+1)^{2}\\ \color{red}\textrm{d}.&4+5+6+...+(k+4)<5(k^{2}+2k+1)\\ \textrm{e}.&4+5+6+...+(k+4)<5(k+1)(k-1) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&4+5+6+...+(n+3)<5n^{2}\\ &\textrm{Saat}\: \: n=k+1,\: \: \textrm{maka}\\ &4+5+6+...+((k+1)+3)<5(k+1)^{2}\\ &=4+5+6+...+(k+4)<5\left ( k^{2}+2k+1 \right ) \end{aligned} \end{array}$