Sebelum kita membahas materi seperti judul di atas, Anda dapat mengulik materi sebelumnya tentang barisan dan deret di link berikut:
Deret Aritmetika dan Geometri Sekaligus
Perhatikan barisan bilangan berikut
$\color{red}\begin{aligned}&a,\: (a+b)r,\: (a+2b)r^{2},\: (a+3b)r^{3},\cdots ,(a+(n-1)b)r^{n-1} \end{aligned}$.
Jika deretnya berhingga maka jumlah deretnya adalah
$\begin{array}{ll} \begin{aligned}S_{n}&=a+\: (a+b)r+\: (a+2b)r^{2}+\cdots +(a+(n-1)b)r^{n-1}\\ rS_{n}&=ar+\: (a+b)r^{2}+\: (a+2b)r^{3}+\cdots +(a+(n-1)b)r^{n}\\ \end{aligned}&- \\\hline \begin{aligned}(1-r)S_{n}&=a-(a+(n-1)b)r^{n}+\displaystyle \frac{br(1-r^{n-1})}{(1-r)}\\ \Leftrightarrow \: \: S_{n}&=\color{red}\displaystyle \frac{a-(a+(n-1)b)r^{n}}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r)^{2}} \end{aligned} \end{array}$.
Jika deretnya tak berhingga, maka nilai $S_{n}$ bergantung pada nilai $\underset{n\rightarrow \color{red}\infty }{\textrm{lim}} \: \displaystyle r^{n}$
- Jika $\color{red}\left | r \right |< 1$, maka deretnya konvergen (memiliki jumlah atau jumlahnya dapat ditentukan), yaitu : $S_{\infty }=\displaystyle \frac{a}{1-r}+\frac{br}{(1-r)^{2}}$ dengan suku awal deret geometrinya adalah 1.
- Jika $\color{red}\left | r \right |\geq 1$, maka deret tak memiliki jumlah yang pas (jumlahnya tidak dapat ditentukan)
Bukti :
untuk $\color{red}\left | r \right |< 1$. Diketahui bahwa $\begin{aligned}S_{n}&=\color{red}\displaystyle \frac{a-(a+(n-1)b)r^{n}}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r)^{2}} \end{aligned}$. Karena harga $\underset{n\rightarrow \color{red}\infty }{\textrm{lim}} \: \displaystyle r^{n}=\color{red}0$, maka jumlah deretnya adalah:
$\begin{aligned}\underset{n\rightarrow \color{red}\infty }{\textrm{lim}} \: \displaystyle S&=\underset{n\rightarrow \color{red}\infty }{\textrm{lim}} \: \color{red}\displaystyle \frac{a-(a+(n-1)b)r^{n}}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r)^{2}}\\ &=\underset{n\rightarrow \color{red}\infty }{\textrm{lim}} \: \color{red}\displaystyle \frac{a-(a+(n-1)b)r^{n}}{(1-r)}+\frac{br-br^{n}}{(1-r)^{2}}\\ &=\underset{n\rightarrow \color{red}\infty }{\textrm{lim}} \: \color{red}\displaystyle \frac{a-\color{black}0}{(1-r)}+\frac{br-\color{black}0}{(1-r)^{2}}\\ &=\underset{n\rightarrow \color{red}\infty }{\textrm{lim}} \: \color{red}\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r)^{2}}\\ &=\color{red}\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r)^{2}}\qquad \color{black}\blacksquare \end{aligned}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah jumlah nilai dari}\\ & \displaystyle \frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots \\\\ &\textbf{Pembahasan}:\\ &\begin{aligned}&S_{\infty }=\displaystyle \frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots \\ &\textrm{dengan}\\ &\begin{array}{|c|c|}\hline \textrm{Bagian aritmetika}&\textrm{Bagian Geometri}\\ (\textrm{lihat bagian pembilang})&(\textrm{bagian pembilang-penyebut})\\\hline \begin{cases} \bullet \quad a& =U_{1}=0 \\ \bullet \: \: \: \: b&=U_{2}-U_{1}\\ &=1-0=1 \end{cases}&\bullet \quad r=\displaystyle \frac{U_{2}}{U_{1}}=\frac{\frac{1}{2}}{1}=\displaystyle \frac{1}{2}\\\hline \end{array}\\ &\begin{aligned}S_{\infty }&=\color{red}\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r)^{2}}\\ &=\displaystyle \frac{0}{1-\frac{1}{2}}+\frac{1\times \frac{1}{2}}{(1-\frac{1}{2})^{2}}=0+\displaystyle \frac{\frac{1}{2}}{\frac{1}{4}}\\ &=0+\displaystyle \frac{4}{2}\\ &=\color{red}2 \end{aligned}\\ &\textrm{Jadi},\: \: \displaystyle \frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots =\color{red}2\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah jumlah nilai dari}\\ & \displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots \\\\ &\textbf{Pembahasan}:\\ &\begin{aligned}&S_{\infty }=\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots \\ &\color{blue}\textrm{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ &2S_{\infty }=1+\displaystyle \frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\frac{9}{16}+\cdots \\ &\begin{array}{|c|c|}\hline \textrm{Bagian aritmetika}&\textrm{Bagian Geometri}\\ (\textrm{lihat bagian pembilang})&(\textrm{bagian pembilang-penyebut})\\\hline \begin{cases} \bullet \quad a& =U_{1}=1 \\ \bullet \: \: \: \: b&=U_{2}-U_{1}=3-1=2 \end{cases}&\bullet \quad r=\displaystyle \frac{U_{2}}{U_{1}}=\frac{\frac{1}{2}}{1}=\displaystyle \frac{1}{2}\\\hline \end{array}\\ &\begin{aligned}2S_{\infty }&=\color{red}\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r)^{2}}\\ 2S_{\infty }&=\displaystyle \frac{1}{1-\frac{1}{2}}+\frac{2\times \frac{1}{2}}{(1-\frac{1}{2})^{2}}=2+4=6\\ S_{\infty }&=\color{red}3 \end{aligned}\\ &\textrm{Jadi},\: \: \displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots =\color{red}3\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah jumlah nilai dari}\\ &2+ \displaystyle \frac{5}{2}+\frac{8}{2^{2}}+\frac{11}{2^{3}}+\frac{14}{2^{4}}+\frac{17}{2^{5}}+\cdots \\\\ &\textbf{Pembahasan}:\\ &\begin{aligned}&S_{\infty }=2+ \displaystyle \frac{5}{2}+\frac{8}{2^{2}}+\frac{11}{2^{3}}+\frac{14}{2^{4}}+\frac{17}{2^{5}}+\cdots\\ &\color{blue}\textrm{dengan}\\ &\begin{array}{|c|c|}\hline \textrm{Bagian aritmetika}&\textrm{Bagian Geometri}\\ (\textrm{lihat bagian pembilang})&(\textrm{bagian pembilang-penyebut})\\\hline \begin{cases} \bullet \quad a& =U_{1}=2 \\ \bullet \: \: \: \: b&=U_{2}-U_{1}=5-2=3 \end{cases}&\bullet \quad r=\displaystyle \frac{U_{2}}{U_{1}}=\frac{\frac{1}{2}}{1}=\displaystyle \frac{1}{2}\\\hline \end{array}\\ &\begin{aligned}S_{\infty }&=\color{red}\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r)^{2}}\\ &=\displaystyle \frac{2}{1-\frac{1}{2}}+\frac{3\times \frac{1}{2}}{(1-\frac{1}{2})^{2}}=4+6=\color{red}10 \end{aligned}\\ &\textrm{Jadi},\: \: 2+ \displaystyle \frac{5}{2}+\frac{8}{2^{2}}+\frac{11}{2^{3}}+\frac{14}{2^{4}}+\frac{17}{2^{5}}+\cdots=\color{red}10\end{aligned} \end{array}$.
$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.
Tentukan besar jumlah dari deret berikut
$\begin{aligned}1.\quad&1+\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\frac{6}{243}+\cdots \\ 2.\quad&1+\displaystyle \frac{3}{3}+\frac{5}{9}+\frac{7}{27}+\frac{9}{81}+\frac{11}{243}+\cdots \\ 3.\quad&1+\displaystyle \frac{2}{5}+\frac{3}{5^{2}}+\frac{4}{5^{3}}+\frac{5}{5^{4}}+\frac{6}{5^{5}}+\cdots \\ 4.\quad&1+\displaystyle \frac{3}{7}+\frac{5}{7^{2}}+\frac{7}{7^{3}}+\frac{9}{7^{4}}+\frac{11}{7^{5}}+\cdots \\ 5.\quad&\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\frac{6}{243}+\frac{7}{729}+\cdots \\ 6.\quad&\displaystyle \frac{3}{3}+\frac{5}{9}+\frac{7}{27}+\frac{9}{81}+\frac{11}{243}+\frac{13}{729}+\cdots \\ 7.\quad&\displaystyle \frac{2}{5}+\frac{3}{5^{2}}+\frac{4}{5^{3}}+\frac{5}{5^{4}}+\frac{6}{5^{5}}+\frac{7}{5^{6}}+\cdots \\ 8.\quad&\displaystyle \frac{3}{7}+\frac{5}{7^{2}}+\frac{7}{7^{3}}+\frac{9}{7^{4}}+\frac{11}{7^{5}}+\frac{13}{7^{6}}+\cdots \\ \end{aligned}$.
DAFTAR PUSTAKA
- Thohir, A. 2013. Barisan dan Deret Materi Pendamping Olimpiade Matematika MA/SMA. Grobogan: MA Futuhiyah.
SUMBER INTERNET
https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence
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