PAS MATEMATIKA BLOG

$\begin{array}{ll} \begin{aligned}S_{n}&=a+\: (a+b)r+\: (a+2b)r^{2}+\cdots +(a+(n-1)b)r^{n-1}\\ rS_{n}&=ar+\: (a+b)r^{2}+\: (a+2b)r^{3}+\cdots +(a+(n-1)b)r^{n}\\ \end{aligned}&- \\\hline \begin{aligned}(1-r)S_{n}&=a-(a+(n-1)b)r^{n}+\displaystyle \frac{br(1-r^{n-1})}{(1-r)}\\ \Leftrightarrow \: \: S_{n}&=\color{red}\displaystyle \frac{a-(a+(n-1)b)r^{n}}{(1-r)}+\frac{br(1-r^{n-1})}{(1-r)^{2}} \end{aligned} \end{array}$.

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

Lanjutan 3 (Barisan & Deret)

$\begin{array}{ll}\\ 11.&\textrm{Hasil penjumlahan dari}\\ &\displaystyle \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\cdots +\frac{1}{97.99}=\cdots \\ &\begin{array}{lllllll}\\ \textrm{A}.&\displaystyle \frac{98}{99}&&&\textrm{D}.&\displaystyle \frac{48}{99}\\\\ \textrm{B}.&\displaystyle \frac{50}{99}\qquad&\textrm{C}.&\color{red}\displaystyle \frac{49}{99}\qquad&\textrm{E}.&\displaystyle \frac{47}{99} \end{array}\\\\ &\textbf{Pembahasan}:\\ &\begin{aligned}&\textrm{Bentuk di atas memenuhi bentuk}\\ &\displaystyle \frac{1}{x(x+2)}=\displaystyle \frac{1}{2}\left ( \displaystyle \frac{1}{x}-\frac{1}{(x+2)} \right ).\: \: \textrm{Bentuk ini pada}\\ &\textrm{bilangan dengan pola tertentu seperti di atas akan}\\ &\textrm{menghabiskan dengan bilangan sebelahnya}\\ &\textrm{atau lazim dikenal dengan}\: \: \textbf{prinsip teleskoping}\\ &\textrm{Sebagaimana bentuk penjumlahan dengan pola di atas}\\ &\textrm{maka}\\ &=\displaystyle \frac{1}{2}\left ( 1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\cdots +\frac{1}{97}-\frac{1}{99} \right )\\ &=\displaystyle \frac{1}{2}\left ( 1-\frac{1}{99} \right )=\displaystyle \frac{1}{2}.\frac{98}{99}=\color{red}\displaystyle \frac{49}{99} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Diketahui}\\ &x=\displaystyle \frac{1+p+p^{2}+p^{3}+\cdots +p^{n-1}}{1+p+p^{2}+p^{3}+\cdots +p^{n-2}+p^{n-1}+p^{n}} \\ &y=\displaystyle \frac{1+q+q^{2}+q^{3}+\cdots +q^{n-1}}{1+q+q^{2}+q^{3}+\cdots +q^{n-2}+q^{n-1}+q^{n}}\\\\ &\textrm{dan}\: \: p>q>0\\\\ &\textrm{Tunjukkan bahwa}\: \: x<y \\\\\\ &\textbf{Bukti}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}:\: \: p>q>0\\ &\textrm{sehingga}\\ &\displaystyle \frac{1}{p}< \frac{1}{q},\: \: \displaystyle \frac{1}{p^{2}}< \frac{1}{q^{2}},\cdots , \displaystyle \frac{1}{p^{n}}< \frac{1}{q^{n}}\\ &\textrm{Jika bentuk di atas dijumlahkan, maka}\\ &\displaystyle \frac{1}{p}+\frac{1}{p^{2}}+\cdots +\frac{1}{p^{n}}< \frac{1}{q}+\frac{1}{q^{2}}+\cdots +\frac{1}{q^{n}}\\ &\Leftrightarrow \displaystyle \frac{p^{n-1}+\cdots +p^{2}+p+1}{p^{n}}< \displaystyle \frac{q^{n-1}+\cdots +q^{2}+q+1}{q^{n}}\\ &\Leftrightarrow \displaystyle \frac{p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}>\displaystyle \frac{q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}\\ &\Leftrightarrow \displaystyle \frac{p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}\color{red}+1\color{black}>\displaystyle \frac{q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}\color{red}+1\\ &\Leftrightarrow \displaystyle \frac{1+p+p^{2}+\cdots +p^{n-1}+p^{n}}{1+p+p^{2}+\cdots +p^{n-1}}>\displaystyle \frac{1+q+q^{2}+\cdots +q^{n-1}+q^{n}}{1+q+q^{2}+\cdots +q^{n-1}}\\ &\Leftrightarrow \displaystyle \frac{1+p+p^{2}+\cdots +p^{n-1}}{1+p+p^{2}+\cdots +p^{n-1}+p^{n}}<\displaystyle \frac{1+q+q^{2}+\cdots +q^{n-1}}{1+q+q^{2}+\cdots +q^{n-1}+q^{n}}\\ &\Leftrightarrow x<y\qquad \blacksquare \end{aligned} \end{array}$.

DAFTAR PUSTAKA

Aziz, A. 2016. Rahasia Juara Olimpiade Matematika SMA. Yogyakarta: ANDI.
Baskoro, B.D. 2012. Aljabar dan Trigonometri Cespleng Olimpiade Matematika. Yogyakarta: BERLIAN.
Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade Nasional & Internasional. Yogyakarta: INDONESIA CERDAS.
Idris, M,. Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
Sembiring, S. 2002. Olimpiade Matematika Untuk SMU. Bandung: YRAMA WIDYA.
Sembiring, S., Suparmin, S. 2015. Pena Emas OSN Matematika SMA. Bandung: YRAMA WIDYA.

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

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

Lanjutan (Barisan & Deret)

$\begin{array}{ll}\\ 5.&\textbf{(Lomba Matematika Nasional}\\ &\textbf{HIMATIKA UGM 2006)} \\ &\textrm{Jika bilangan}\\ &A=\displaystyle \frac{1}{1+1}+\frac{1}{1+2}+\frac{1}{1+3}+\cdots +\frac{1}{1+100}\\ &B=\displaystyle \frac{1}{1+1}+\frac{1}{1+\displaystyle \frac{1}{2}}+\frac{1}{1+\displaystyle \frac{1}{3}}+\cdots +\frac{1}{1+\displaystyle \frac{1}{100}}\\ &\textrm{maka}\: \: A+B\: \: \textrm{sama dengan}\: ....\\ &\textrm{A}.\quad 202 \: \: \qquad\qquad\qquad\qquad\qquad \textrm{D}.\quad \color{red}100\\ &\textrm{B}.\quad 200\qquad\qquad \color{black}\textrm{C}.\quad 101\qquad\quad \color{black}\textrm{E}.\quad 99\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\\ &\begin{array}{ll}\\ \begin{aligned}A&=\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{101}\\ B&=\displaystyle \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots +\frac{100}{101}\\ \end{aligned}&\\&+\\\hline \\ A+B=\underset{100}{\underbrace{1+1+1+\cdots +1}}&=\color{red}100 \end{array} \end{array}$.

$\begin{array}{ll}\\ 6.&\textbf{(OSN tk. Kota/Kab 2002)} \\ &\textrm{Misalkan}\\ &a=\displaystyle \frac{1^{2}}{1}+\frac{2^{2}}{3}+\frac{3^{2}}{5}+\cdots +\frac{1001^{2}}{2001}\\\\ &b=\displaystyle \frac{1^{2}}{3}+\frac{2^{2}}{5}+\frac{3^{2}}{7}+\cdots +\frac{1001^{2}}{2003}\\ &\textrm{Tentukanlah bilangan bulat yang}\\ &\textrm{nilainya paling dekat ke}\: \: a-b\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}a-b &=\left (\displaystyle \frac{1^{2}}{1}+\frac{2^{2}}{3}+\frac{3^{2}}{5}+\cdots +\frac{1001^{2}}{2001} \right )\\ &\: -\left ( \displaystyle \frac{1^{2}}{3}+\frac{2^{2}}{5}+\frac{3^{2}}{7}+\cdots +\frac{1001^{2}}{2003} \right )\\ &=\displaystyle \frac{1^{2}}{1}+\left ( \displaystyle \frac{2^{2}}{3}-\displaystyle \frac{1^{2}}{3} \right )+\left ( \displaystyle \frac{3^{2}}{5}-\displaystyle \frac{2^{2}}{5} \right )\\ &\: +\left ( \displaystyle \frac{4^{2}}{7}-\displaystyle \frac{3^{2}}{7} \right )+\cdots +\left ( \displaystyle \frac{1001^{2}}{2001}-\displaystyle \frac{1000^{2}}{2001} \right )\\ &\: \: \: -\displaystyle \frac{1001^{2}}{2003}\\ &=\underset{1001}{\underbrace{1+1+1+\cdots +1}}-\displaystyle \frac{1001^{2}}{2003}\\ &=1001-\displaystyle \frac{1001^{2}}{2003}=1001\left ( \displaystyle \frac{2003-1001}{2003} \right )\\ &=\displaystyle \frac{1001\times 1002}{2003}> \displaystyle \frac{1001}{2}=\color{red}500,5 \end{aligned}\\ &\textrm{Jadi bilangan bulat yang paling dekat}\\ &\textrm{ke}\: \: a-b\: \: \textrm{adalah}\: \: \color{red}501 \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Misalkan}\\ &a_{n}=\displaystyle \frac{n(n+1)}{2},\: \textrm{tentukanlah jumlah}\\ &\textrm{dari}\: \: \displaystyle \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots +\frac{1}{a_{2023}}\\\\ &\textbf{Pembahasan}:\\ &\begin{aligned}a_{n}&=\displaystyle \frac{n(n+1)}{2},\: \: \textrm{maka}\\ \displaystyle \frac{1}{a_{n}}&=\displaystyle \frac{2}{n(n+1)}\\ &=2\left ( \displaystyle \frac{1}{n}-\frac{1}{n+1} \right )\\ &\color{red}\textrm{lihat pembahasan no.3 di atas}\\ & \end{aligned} \\ &\begin{aligned}&\displaystyle \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots +\frac{1}{a_{2023}}\\ &=2\left ( \left (1-\displaystyle \frac{1}{2} \right )+\left ( \displaystyle \frac{1}{2}-\frac{1}{3} \right )+\cdots +\left ( \displaystyle \frac{1}{2022}-\frac{1}{2023} \right ) \right )\\ &=2\left ( 1-\displaystyle \frac{1}{2023} \right )\\ &=2\left (\displaystyle \frac{2022}{2023} \right )=\color{blue}\displaystyle \frac{4044}{2023} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Misalkan}\: \: n\: \: \textrm{adalah bilangan asli}\\ &\textrm{dan}\: \: \left \{ a_{n} \right \}\: \textrm{adalah barisan bilangan real}\\ &\textrm{dengan}\: \: a_{n}=\displaystyle \frac{2^{n}}{2^{2n+1}-2^{n+1}-2^{n}+1}\\\\ &\textrm{Tunjukkan bahwa untuk setiap bilangan}\\ &\textrm{asli}\: \: n, \: \: \textrm{berlaku}\: \: \: a_{1}+a_{2}+\cdots +a_{n}<1\\\\ &\textbf{Bukti}:\\ &\begin{aligned}a_{n}&=\displaystyle \frac{2^{n}}{2^{2n+1}-2^{n+1}-2^{n}+1}\\ &=\displaystyle \frac{2^{n}}{(2^{n+1}-1)(2^{n}-1)}\\ &=\left ( \displaystyle \frac{1}{2^{n+1}-1}-\frac{1}{2^{n}-1} \right )\\ a_{1}&=\displaystyle \frac{1}{2^{1}-1}-\frac{1}{2^{2}-1}=1-\displaystyle \frac{1}{3}\\ a_{2}&=\displaystyle \frac{1}{2^{2}-1}-\frac{1}{2^{3}-1}=\frac{1}{3}-\frac{1}{7}\\ a_{3}&=\displaystyle \frac{1}{2^{3}-1}-\frac{1}{2^{4}-1}=\frac{1}{7}-\frac{1}{15}\\ &\vdots \qquad\qquad\qquad \vdots \\ a_{n}&=\displaystyle \frac{1}{2^{n}-1}-\frac{1}{2^{n+1}-1}\quad\quad\quad\quad +\\ \color{purple}a_{1}&\color{purple}+a_{2}+a_{3}+\cdots +a_{n}\\ &=\color{red}1-\displaystyle \frac{1}{2^{n+1}-1}< 1\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

Selingan (Barisan & Deret)

Lanjutan contoh soal dan pembahasannya terkait barisan dan deret

$\begin{array}{ll}\\ 1.&\textrm{Hasil kali bilangan bentuk berikut}\\ &\left ( 1-\displaystyle \frac{1}{4} \right )\left ( 1-\displaystyle \frac{1}{5} \right )\left ( 1-\displaystyle \frac{1}{6} \right )\cdots \left ( 1-\displaystyle \frac{1}{100} \right )\\ &\textrm{adalah}\: ....\\ &\textrm{A}.\quad \displaystyle \frac{1}{100} \: \: \qquad\qquad\qquad\qquad\qquad \textrm{D}.\quad \displaystyle \frac{4}{100}\\\\ &\textrm{B}.\quad \displaystyle \frac{2}{100}\qquad\qquad \color{black}\textrm{C}.\quad \displaystyle \color{red}\frac{3}{100}\qquad\quad \color{black}\textrm{E}.\quad \displaystyle \frac{5}{100}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\\ &\begin{aligned}&\left ( 1-\displaystyle \frac{1}{4} \right )\left ( 1-\displaystyle \frac{1}{5} \right )\left ( 1-\displaystyle \frac{1}{6} \right )\cdots \left ( 1-\displaystyle \frac{1}{100} \right )\\ &=\left ( \displaystyle \frac{3}{4} \right )\left ( \displaystyle \frac{4}{5} \right )\left ( \displaystyle \frac{5}{6} \right )\cdots \left ( \displaystyle \frac{99}{100} \right )\\ &=\left ( \displaystyle \frac{\color{red}3}{\not{4}} \right )\left ( \displaystyle \frac{\not{4}}{\not{5}} \right )\left ( \displaystyle \frac{\not{5}}{\not{6}} \right )\cdots \left ( \displaystyle \frac{\not{99}}{\color{red}100} \right )\\ &=\color{red}\displaystyle \frac{3}{100} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Hasil kali bilangan bentuk berikut}\\ &\left ( 1-\displaystyle \frac{2}{3} \right )\left ( 1-\displaystyle \frac{2}{5} \right )\left ( 1-\displaystyle \frac{2}{7} \right )\cdots \left ( 1-\displaystyle \frac{2}{2023} \right )\\ &\textrm{adalah}\: ....\\ &\textrm{A}.\quad \color{red}\displaystyle \frac{1}{2023} \: \: \qquad\qquad\qquad\qquad\qquad\: \: \textrm{D}.\quad \displaystyle \frac{4}{2023}\\\\ &\textrm{B}.\quad \displaystyle \frac{2}{2023}\qquad\qquad \color{black}\textrm{C}.\quad \displaystyle \frac{3}{2023}\qquad\quad \color{black}\textrm{E}.\quad \displaystyle \frac{5}{2023}\\\\ &\textbf{Jawab}:\\ &\textrm{Dengan cara pembahasan pada no.1 di atas, maka}\\ &\begin{aligned}&\left ( 1-\displaystyle \frac{2}{3} \right )\left ( 1-\displaystyle \frac{2}{5} \right )\left ( 1-\displaystyle \frac{2}{7} \right )\cdots \left ( 1-\displaystyle \frac{2}{2023} \right )\\ &=\left ( \displaystyle \frac{1}{3} \right )\left ( \displaystyle \frac{3}{5} \right )\left ( \displaystyle \frac{5}{7} \right )\cdots \left ( \displaystyle \frac{2021}{2023} \right )\\ &=\left ( \displaystyle \frac{\color{red}1}{\not{3}} \right )\left ( \displaystyle \frac{\not{3}}{\not{5}} \right )\left ( \displaystyle \frac{\not{5}}{\not{7}} \right )\cdots \left ( \displaystyle \frac{\not{2021}}{\color{red}2023} \right )\\ &=\color{red}\displaystyle \frac{1}{2023} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Bentuk sederhana dari}\\ &\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{2022\times 2023}\\\\ &\textbf{Pembahasan}:\\ &\begin{aligned}&=\left ( 1-\displaystyle \frac{1}{2} \right )+\left ( \displaystyle \frac{1}{2}-\frac{1}{3} \right )+\left ( \displaystyle \frac{1}{3}-\frac{1}{4} \right )+\\ & \qquad\qquad +\quad\cdots\qquad +\left ( \displaystyle \frac{1}{2022}-\frac{1}{2023} \right )\\ &=1-\displaystyle \frac{1}{2023}\\ &=\color{blue}\displaystyle \frac{2022}{2023} \end{aligned} \end{array}$.

$\begin{array}{|c|}\hline \begin{aligned}\color{blue}\textbf{Seb}&\color{blue}\textbf{agai pengingat kita}\\ \bullet \quad&1+2+3+\cdots +n=\displaystyle \frac{n(n+1)}{2}\\ \bullet \quad&1+3+5+\cdots +(2n-1)=n^{2}\\ \bullet \quad&1^{2}+2^{2}+3^{2}+\cdots +n^{2}=\displaystyle \frac{n(n+1)(2n+1)}{6}\\ \bullet \quad&1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left ( \displaystyle \frac{n(n+1)}{2} \right )^{2}\\ \bullet \quad&1+\displaystyle \frac{1}{1+2}+\frac{1}{1+2+3}+\cdots +\frac{1}{1+2+3+\cdots +n}=\displaystyle \frac{2n}{n+1}\\ \bullet \quad&\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{n(n+1)}=\displaystyle \frac{n}{n+1}\\ \bullet \quad&\displaystyle \frac{1}{1.2.3}+\frac{1}{2.3.4}+\cdots +\frac{1}{n(n+1)(n+2)}=\displaystyle \frac{n(n+3)}{4(n+1)(n+2)} \end{aligned}\\\hline \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Bentuk sederhana dari}\\ &\displaystyle \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{2023}-\sqrt{2022}}\\\\ &\textbf{Pembahasan}:\\ &\begin{aligned}\displaystyle \frac{1}{1+\sqrt{2}}&=\frac{1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}\\ &=\displaystyle \frac{1-\sqrt{2}}{1^{2}-(\sqrt{2})^{2}}=\frac{1-\sqrt{2}}{1-2}\\ &=\displaystyle \frac{1-\sqrt{2}}{-1}=\color{red}\sqrt{2}-1\\ \displaystyle \frac{1}{\sqrt{2}+\sqrt{3}}&=\frac{1}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\\ &=\displaystyle \frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}=\frac{\sqrt{2}-\sqrt{3}}{2-3}\\ &=\displaystyle \frac{\sqrt{2}-\sqrt{3}}{-1}=\color{red}\sqrt{3}-\sqrt{2}\\ \displaystyle \frac{1}{\sqrt{3}+\sqrt{4}}&=\frac{1}{\sqrt{3}+\sqrt{4}}\times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\\ &=\displaystyle \frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3})^{2}-(\sqrt{4})^{2}}=\frac{\sqrt{3}-\sqrt{4}}{3-4}\\ &=\displaystyle \frac{\sqrt{3}-\sqrt{4}}{-1}=\color{red}\sqrt{4}-\sqrt{3}\\ &\vdots \\ \end{aligned}\\ &\textrm{Sehingga}\\ &\begin{aligned}&=\color{red}\not{\sqrt{2}}-1\color{black}+\color{red}\not{\sqrt{3}}-\not{\sqrt{2}}\color{black}+\color{red}\not{\sqrt{4}}-\not{\sqrt{3}}\color{black}+\color{red}\cdots \color{black}+\color{red}\sqrt{2023}-\not{\sqrt{2022}}\\ &=\color{blue}\sqrt{2023}-1 \end{aligned} \end{array}$