Lanjutan 2 (Barisan & Deret)

$\begin{array}{ll}\\ 9.&\textrm{Tentukan hasil dari}\\ &\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots \\\\ &\textbf{Pembahasan}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\begin{aligned}&\textrm{Misalkan}\: \: S=\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots \\ &\textrm{maka}\: \: \displaystyle \frac{1}{2}S=\displaystyle \frac{1}{4}+\frac{3}{8}+\frac{5}{16}+\frac{7}{32}+\frac{9}{64}+\cdots\\ &\textrm{Jika}\\ &S-\displaystyle \frac{1}{2}S=\displaystyle \frac{1}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2}{64}+\cdots \\ &\begin{aligned}\Leftrightarrow \displaystyle \frac{1}{2}S&=\displaystyle \frac{1}{2}+\underset{\textrm{deret geometri}}{\underbrace{\left (\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots   \right )}}\\ \Leftrightarrow \displaystyle \frac{1}{2}S&=\displaystyle \frac{1}{2}+\displaystyle \frac{\displaystyle \frac{1}{2}}{1-\displaystyle \frac{1}{2}}=\displaystyle \frac{1}{2}+\displaystyle \frac{\frac{1}{2}}{\frac{1}{2}}=\displaystyle \frac{1}{2}+1=\color{blue}\displaystyle \frac{3}{2}\\ \Leftrightarrow \displaystyle \frac{1}{2}S&=\displaystyle \frac{3}{2}\\ \Leftrightarrow \, \: \: \:  S&=\color{red}3 \end{aligned} \end{aligned}\\ &\color{blue}\textbf{Alternatif 2}\\ &\begin{aligned}&S_{\infty }=\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots \\ &\color{purple}\textrm{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ &2S_{\infty }=1+\displaystyle \frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\frac{9}{16}+\cdots \\ &\begin{array}{|c|c|}\hline \textrm{Bagian aritmetika}&\textrm{Bagian Geometri}\\ (\textrm{lihat bagian pembilang})&(\textrm{bagian pembilang-penyebut})\\\hline \begin{cases} \bullet  \quad a& =U_{1}=1 \\  \bullet  \: \: \: \: b&=U_{2}-U_{1}=3-1=2  \end{cases}&\bullet \quad r=\displaystyle \frac{U_{2}}{U_{1}}=\frac{\frac{1}{2}}{1}=\displaystyle \frac{1}{2}\\\hline \end{array}\\ &\begin{aligned}2S_{\infty }&=\color{red}\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r)^{2}}\\ 2S_{\infty }&=\displaystyle \frac{1}{1-\frac{1}{2}}+\frac{2\times \frac{1}{2}}{(1-\frac{1}{2})^{2}}=2+4=6\\ S_{\infty }&=\color{red}3 \end{aligned}\\ &\textrm{Jadi},\: \:   \displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots  =\color{red}3\end{aligned} \end{array}$.

$.\qquad  \textrm{Untuk link materi pada pembahasan alternatif 2}$. di sini

$\begin{array}{ll}\\ 10.&\textrm{Hasil penjumlahan dari}\\ &\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots =\cdots \\ &\begin{array}{lllllll}\\ \textrm{A}.&\displaystyle \frac{2}{3}&&&\textrm{D}.&\color{red}\displaystyle \frac{4}{3}\\\\ \textrm{B}.&\displaystyle \frac{3}{4}\qquad&\textrm{C}.&1\qquad&\textrm{E}.&\displaystyle \frac{3}{2} \end{array}\\\\ &\textbf{Pembahasan}:\\ &\begin{aligned}&S_{\infty }=\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots \\ &\color{purple}\textrm{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ &3S_{\infty }=1+\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\cdots \\ &\begin{array}{|c|c|}\hline \textrm{Bagian aritmetika}&\textrm{Bagian Geometri}\\ (\textrm{lihat bagian pembilang})&(\textrm{bagian pembilang-penyebut})\\\hline \begin{cases} \bullet  \quad a& =U_{1}=1 \\  \bullet  \: \: \: \: b&=U_{2}-U_{1}=2-1=1  \end{cases}&\bullet \quad r=\displaystyle \frac{U_{2}}{U_{1}}=\frac{\frac{1}{3}}{1}=\displaystyle \frac{1}{3}\\\hline \end{array}\\ &\begin{aligned}3S_{\infty }&=\color{red}\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r)^{2}}\\ 3S_{\infty }&=\displaystyle \frac{1}{1-\frac{1}{3}}+\frac{1\times \frac{1}{3}}{(1-\frac{1}{3})^{2}}=\displaystyle \frac{3}{2}+\frac{3}{4}=\displaystyle \frac{9}{4}\\ S_{\infty }&=\color{red}\displaystyle \frac{3}{4} \end{aligned}\\ &\textrm{Jadi},\: \:   \displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots =\color{red}\displaystyle \frac{3}{4}\end{aligned} \end{array}$ .