Lanjutan 2 Materi Barisan dan Deret (Deret Aritmetika-Geometri tak Berhingga)

Sebelum kita membahas materi seperti judul di atas, Anda dapat mengulik materi sebelumnya tentang barisan dan deret di link berikut:

• link 1 (Barisan dan Deret)
• link 2 (Barisan dan Deret)

Deret Aritmetika dan Geometri Sekaligus

Perhatikan barisan bilangan berikut

$\begin{array}{rl}& a,(a+b)r,(a+2b)r^2,(a+3b)r^3,\cdots ,(a+(n-1)b)r^{n-1}\end{array}$.

Jika deretnya berhingga maka jumlah deretnya adalah

$\begin{array}{ll}\begin{array}{rl}{S}_n& =a+(a+b)r+(a+2b)r^2+\cdots +(a+(n-1)b)r^{n-1}\\ r{S}_n& =ar+(a+b)r^2+(a+2b)r^3+\cdots +(a+(n-1)b)r^n\end{array}& -\\ \begin{array}{rl}(1-r)S_n& =a-(a+(n-1)b)r^n+{\displaystyle \frac{br(1-r^{n-1})}{(1-r)}}\\ \iff S_n& ={\displaystyle \frac{a-(a+(n-1)b)r^n}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r{)}^2}}\end{array}\end{array}$.

Jika deretnya tak berhingga, maka nilai  $S_n$  bergantung pada nilai  $\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle r^n}$

• Jika  $\left|r\right|<1$, maka deretnya konvergen (memiliki jumlah atau jumlahnya dapat ditentukan), yaitu : $S_{\mathrm{\infty }}={\displaystyle \frac{a}{1-r}+\frac{br}{(1-r{)}^2}}$ dengan suku awal deret geometrinya adalah 1.
• Jika  $\left|r\right|\ge 1$, maka deret tak memiliki jumlah yang pas (jumlahnya tidak dapat ditentukan)

Bukti :

untuk $\left|r\right|<1$. Diketahui bahwa  $\begin{array}{rl}{S}_n& ={\displaystyle \frac{a-(a+(n-1)b)r^n}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r{)}^2}}\end{array}$. Karena harga  $\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle r^n=0}$, maka jumlah deretnya adalah:

$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle S}& =\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{a-(a+(n-1)b)r^n}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r{)}^2}}\\ & =\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{a-(a+(n-1)b)r^n}{(1-r)}+\frac{br-b{r}^n}{(1-r{)}^2}}\\ & =\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{a-0}{(1-r)}+\frac{br-0}{(1-r{)}^2}}\\ & =\underset{n\to \mathrm{\infty }}{\text{lim}}{\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ & ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}◼}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah jumlah nilai dari}\\ & {\displaystyle \frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots }\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& S_{\mathrm{\infty }}={\displaystyle \frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots }\\ & \text{dengan}\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=0\\ \bullet b& =U_2-U_1\\ & =1-0=1\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{2}}{1}={\displaystyle \frac{1}{2}}}\\ \hline\end{array}\\ & \begin{array}{rl}{S}_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ & ={\displaystyle \frac{0}{1-\frac{1}{2}}+\frac{1\times \frac{1}{2}}{(1-\frac{1}{2}{)}^2}=0+{\displaystyle \frac{\frac{1}{2}}{\frac{1}{4}}}}\\ & =0+{\displaystyle \frac{4}{2}}\\ & =2\end{array}\\ & \text{Jadi},{\displaystyle \frac{0}{1}+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\cdots =2}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 3.& \text{Tentukanlah jumlah nilai dari}\\ & 2+{\displaystyle \frac{5}{2}+\frac{8}{2^2}+\frac{11}{2^3}+\frac{14}{2^4}+\frac{17}{2^5}+\cdots }\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& S_{\mathrm{\infty }}=2+{\displaystyle \frac{5}{2}+\frac{8}{2^2}+\frac{11}{2^3}+\frac{14}{2^4}+\frac{17}{2^5}+\cdots }\\ & \text{dengan}\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=2\\ \bullet b& =U_2-U_1=5-2=3\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{2}}{1}={\displaystyle \frac{1}{2}}}\\ \hline\end{array}\\ & \begin{array}{rl}{S}_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ & ={\displaystyle \frac{2}{1-\frac{1}{2}}+\frac{3\times \frac{1}{2}}{(1-\frac{1}{2}{)}^2}=4+6=10}\end{array}\\ & \text{Jadi},2+{\displaystyle \frac{5}{2}+\frac{8}{2^2}+\frac{11}{2^3}+\frac{14}{2^4}+\frac{17}{2^5}+\cdots =10}\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

Tentukan besar jumlah dari deret berikut

$\begin{array}{rl}1.& 1+{\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\frac{6}{243}+\cdots }\\ 2.& 1+{\displaystyle \frac{3}{3}+\frac{5}{9}+\frac{7}{27}+\frac{9}{81}+\frac{11}{243}+\cdots }\\ 3.& 1+{\displaystyle \frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+\frac{5}{5^4}+\frac{6}{5^5}+\cdots }\\ 4.& 1+{\displaystyle \frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}+\frac{11}{7^5}+\cdots }\\ 5.& {\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\frac{6}{243}+\frac{7}{729}+\cdots }\\ 6.& {\displaystyle \frac{3}{3}+\frac{5}{9}+\frac{7}{27}+\frac{9}{81}+\frac{11}{243}+\frac{13}{729}+\cdots }\\ 7.& {\displaystyle \frac{2}{5}+\frac{3}{5^2}+\frac{4}{5^3}+\frac{5}{5^4}+\frac{6}{5^5}+\frac{7}{5^6}+\cdots }\\ 8.& {\displaystyle \frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}+\frac{11}{7^5}+\frac{13}{7^6}+\cdots }\end{array}$.

DAFTAR PUSTAKA

1. 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