Lanjutan 2 (Barisan & Deret)

$\begin{array}{l}\\ 9.& \text{Tentukan hasil dari}\\ & {\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}& \text{Misalkan}S={\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ & \text{maka}{\displaystyle \frac{1}{2}S={\displaystyle \frac{1}{4}+\frac{3}{8}+\frac{5}{16}+\frac{7}{32}+\frac{9}{64}+\cdots }}\\ & \text{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{array}{rl}\iff {\displaystyle \frac{1}{2}S}& ={\displaystyle \frac{1}{2}+\underset{\text{deret geometri}}{\underbrace{(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots )}}}\\ \iff {\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={\displaystyle \frac{3}{2}}}}}}}\\ \iff {\displaystyle \frac{1}{2}S}& ={\displaystyle \frac{3}{2}}\\ \iff S& =3\end{array}\end{array}\\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& S_{\mathrm{\infty }}={\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ & \text{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ & 2S_{\mathrm{\infty }}=1+{\displaystyle \frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\frac{9}{16}+\cdots }\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=1\\ \bullet b& =U_2-U_1=3-1=2\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{2}}{1}={\displaystyle \frac{1}{2}}}\\ \hline\end{array}\\ & \begin{array}{rl}2S_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ 2S_{\mathrm{\infty }}& ={\displaystyle \frac{1}{1-\frac{1}{2}}+\frac{2\times \frac{1}{2}}{(1-\frac{1}{2}{)}^2}=2+4=6}\\ S_{\mathrm{\infty }}& =3\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots =3}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 10.& \text{Hasil penjumlahan dari}\\ & {\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots =\cdots }\\ & \begin{array}{l}\\ \text{A}.& {\displaystyle \frac{2}{3}}& & & \text{D}.& {\displaystyle \frac{4}{3}}\\ \\ \text{B}.& {\displaystyle \frac{3}{4}}& \text{C}.& 1& \text{E}.& {\displaystyle \frac{3}{2}}\end{array}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& S_{\mathrm{\infty }}={\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots }\\ & \text{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ & 3S_{\mathrm{\infty }}=1+{\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\cdots }\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=1\\ \bullet b& =U_2-U_1=2-1=1\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{3}}{1}={\displaystyle \frac{1}{3}}}\\ \hline\end{array}\\ & \begin{array}{rl}3S_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ 3S_{\mathrm{\infty }}& ={\displaystyle \frac{1}{1-\frac{1}{3}}+\frac{1\times \frac{1}{3}}{(1-\frac{1}{3}{)}^2}={\displaystyle \frac{3}{2}+\frac{3}{4}={\displaystyle \frac{9}{4}}}}\\ S_{\mathrm{\infty }}& ={\displaystyle \frac{3}{4}}\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots ={\displaystyle \frac{3}{4}}}\end{array}\end{array}$ .