$\begin{array}{ll}\\ 31.&2+4+6+8+\cdots =\: ....\:.\\ &\textrm{A}.\quad \color{red}\displaystyle \sum_{k=1}^{n}2k\\ &\textrm{B}.\quad \displaystyle \sum_{k=1}^{n}2^{k}\\ &\textrm{C}.\quad \displaystyle \sum_{k=1}^{n}2k-1\\ &\textrm{D}.\quad \displaystyle \sum_{k=1}^{n}(2-k)\\ &\textrm{E}.\quad \displaystyle \sum_{k=1}^{n}k\\\\ &\textbf{Jawab}:\\ &\textrm{Cukup Jelas bahwa}\\ &\begin{aligned}2+4+6+8+\cdots \: \: \quad&\\ 2(1+2+3+4+\cdots )&=2\displaystyle \sum_{k=1}^{n}k=\displaystyle \sum_{k=1}^{n}\color{red}2k \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 32.&\textbf{(UN 2005)}\\ &\textrm{Seorang anak menabung di suatu bank dengan}\\ &\textrm{selisih kenaikan tabungan antarbulan tetap}\\ &\textrm{Pada bulan pertama sebesar Rp50.000,00},\\ &\textrm{bulan kedua Rp55.000,00, bulan ketiga}\\ &\textrm{Rp60.000,00, dan demikian seterusnya}.\\ &\textrm{Besar tabungan anak tersebut selama }\\ &\textrm{dua tahun adalah}\: ....\:.\\ &\textrm{A}.\quad \textrm{Rp1.315.000,00}\\ &\textrm{B}.\quad \textrm{Rp1.320.000,00}\\ &\textrm{C}.\quad \textrm{Rp2.040.000,00}\\ &\textrm{D}.\quad \color{red}\textrm{Rp2.580.000,00}\\ &\textrm{E}.\quad \textrm{Rp2.640.000,00}\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui deret aritmetika dengan}\\ &\begin{aligned}&\bullet \quad a=U_{1}=\textrm{Rp}50.000,00\\ &\bullet \quad U_{2}=\textrm{Rp}55.000,00\\ &\bullet \quad b=U_{2}-U_{1}=\textrm{Rp}5.000,00\\ &\textrm{Ditanya: Besar tabungan selama 2 tahun}\\ &\begin{aligned}\color{blue}S_{n}&=\color{blue}\displaystyle \frac{n}{2}\left ( 2a+(n-1)b \right )\\ \textrm{Ka}&\textrm{rena 2 tahun = 24 bulan, maka}\\ S_{24}&=\displaystyle \frac{24}{2}(2\times 50.000+(24-1)\times 5.000)\\ &=12\left ( 100.000+115.000 \right )=\color{red}2.580.000 \end{aligned} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 33.&\textbf{(UN 2006)}\\ &\textrm{Sebuah bola jatuh dari ketinggian 10 meter dan}\\ &\textrm{memantul kembali dengan ketinggian}\: \: \displaystyle \frac{3}{4}\: \: \textrm{dari}\\ &\textrm{tinggi semula dan begitu seterusnya hingga bola}\\ &\textrm{berhenti. Jumlah seluruh lintasan bola adalah}\: ....\:.\\ &\textrm{A}.\quad 65\: \: \textrm{meter}\\ &\textrm{B}.\quad \color{red}70\: \: \textrm{meter}\\ &\textrm{C}.\quad 75\: \: \textrm{meter}\\ &\textrm{D}.\quad 77\: \: \textrm{meter}\\ &\textrm{E}.\quad 80\: \: \textrm{meter}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi gambar berikut} \end{array}$.
$.\qquad\begin{aligned}&\textrm{Soal terkait dengan deret geometri tak hingga}\\ &\textrm{Yaitu}:\: S_{\infty }=\displaystyle \frac{a}{1-r},\: \: \textrm{dengan}\: \: \left | r \right |<1\\ &\color{blue}\begin{aligned}S&=10+2.\displaystyle \frac{3}{4}.10+2.\frac{3}{4}.\frac{3}{4}.10+2.\frac{3}{4}.\frac{3}{4}.\frac{3}{4}.10+...\\ &=10+20.\displaystyle \frac{3}{4}+20.\left (\frac{3}{4} \right )^{2}+20+\left ( \displaystyle \frac{3}{4} \right )^{3}+...\\ &=10+20\left ( \displaystyle \frac{3}{4}+\frac{9}{16}+\frac{27}{64}+... \right ) \end{aligned}\\ &\textrm{adalah deret geometri tak hingga dengan}\\ &a=r=\displaystyle \frac{3}{4},\: \: \textrm{maka}\\ &\begin{aligned}S_{\infty }&=10+20.\left (\displaystyle \frac{\displaystyle \frac{3}{4}}{1-\displaystyle \frac{3}{4}} \right )=10+20.\left ( \displaystyle \frac{\displaystyle \frac{3}{4}}{\displaystyle \frac{1}{4}} \right )\\ &=10+20.2=10+60=\color{red}70\: \: \textrm{meter} \end{aligned} \end{aligned}$.
$.\qquad\begin{aligned}&\textbf{Alternatif Jawaban}\\ &\textrm{Dengan rumus praktis, yaitu}\\ &\begin{array}{|l|}\hline\\ \begin{aligned}&\color{blue}\textrm{Panjang seluruh lintasan bola}\\ & \end{aligned}\\ \begin{array}{|l|}\hline \\\begin{aligned}S&=\textrm{Jatuh 1}\times \displaystyle \frac{\textrm{Jumlah perbandingan}}{\textrm{Selisih perbandingan}}\\ &=10\times \displaystyle \frac{4+3}{4-3}=10\times \displaystyle \frac{7}{1}=\color{red}70\: \: \textrm{meter}\\\ \end{aligned}\\\hline \end{array}\\ \\\hline \end{array} \end{aligned}$.
$\begin{array}{ll}\\ 34.&^{4}\log 2+\: ^{4}\log 4+\: ^{4}\log 16+^{4}\log 64+...\\ &\textrm{membentuk}\: ....\:.\\ &\textrm{A}.\quad \textrm{deret aritmetika dengan beda}\: \: ^{4}\log 2\\ &\textrm{B}.\quad \textrm{deret geometri dengan pembanding}\: \: ^{4}\log 2\\ &\textrm{C}.\quad \textrm{deret aritmetika dengan beda 2}\\ &\textrm{D}.\quad \textrm{deret geometri dengan pembanding 2}\\ &\textrm{E}.\quad \color{red}\textrm{bukan deret geometri maupun matematika}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}S_{n}&=\: ^{4}\log 2+\: ^{4}\log 4+\: ^{4}\log 16+^{4}\log 64+...\\ &=\: ^{4}\log 4^{\frac{1}{2}}+\: ^{4}\log 4^{1}+\: ^{4}\log 4^{2}+^{4}\log 4^{3}+...\\ &=\displaystyle \frac{1}{2}+1+2+3+...\\ &\textrm{dengan}\quad a=U_{1}=\displaystyle \frac{1}{2},\: U_{2}=1,\: \&\: U_{3}=2\\ &\textrm{Kita perlu cek dengan ciri masing-masing}\\ &\textrm{deret, yaitu}:\\ &\begin{array}{|l|l|}\hline \textrm{Deret Aritmetika}&\textrm{Deret Geometri}\\\hline \begin{aligned}&2U_{2}=U_{1}+U_{3}\\ &\qquad\color{blue}\textrm{atau}\\ &2U_{n+1}=U_{n}+U_{n+2} \end{aligned}&\begin{aligned}&U_{2}^{2}=U_{1}\times U_{3}\\ &\qquad\color{blue}\textrm{atau}\\ &U_{n+1}^{2}=U_{n}\times U_{n+2} \end{aligned}\\\hline \begin{aligned}&\color{red}2.(1)\neq \displaystyle \frac{1}{2}+2\\ & \end{aligned}&\begin{aligned}&(1)^{2}= \displaystyle \frac{1}{2}\times 2\\ &\color{red}(2)^{2}\neq 1\times 2 \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 35.&\textrm{Suatu modal sebesar}\: \: M\: \: \textrm{rupiah dibungakan dengan}\\ &\textrm{bunga}\: \: p\%\: \: \textrm{pertahun. Jika bunganya majmuk, maka}\\ &\textrm{setelah}\: \: n\: \: \textrm{tahun modal tersebut akan menjadi}\: ...\: .\\ &\textrm{A}.\quad M+(p/100)^{n}\\ &\textrm{B}.\quad (M+p\%.M)^{n}\\ &\textrm{C}.\quad nM.p\%\\ &\textrm{D}.\quad M(1-0,5)^{n}\\ &\textrm{E}.\quad \color{red}M(1+p\%)^{n}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Untuk kasus bunga majmuk di atas adalah}:\\ &\color{red}M,M(1+p\%),M(1+p\%)^{2},M(1+p\%)^{3},\cdots \\ &\textrm{adalah barisan geometri}\\ &\color{red}M_{n}=M_{0}(1+p\%)^{n}\: \: \color{black}\textrm{atau}\: \: M_{n}=M_{0}(1+i)^{n}\\ &\textrm{dengan}\\ &\circ \quad i=p\%\: \: \: \textrm{adalah persentase bunga}\\ &\circ \quad n =\textrm{Jangka waktu}\\ &\circ \quad M_{0}=\textrm{Modal yang diperbungakan} \end{aligned} \end{array}$.
Tidak ada komentar:
Posting Komentar
Informasi