CONTOH SOAL 3 BARISAN

 $\begin{aligned}5.\quad&(\textbf{KSM Matematika MA Tk. Kab/kota Th.2013})\\ &\textrm{Hasil penjumlahan dari tak hingga suku berbentuk}\\ &\displaystyle \frac{1}{5}+\frac{2}{5^{\displaystyle 2}}+\frac{1}{5^{\displaystyle 3}}+\frac{2}{5^{\displaystyle 4}}+\frac{1}{5^{\displaystyle 5}}+\frac{2}{5^{\displaystyle 6}}\:+\:...\quad \textrm{adalah}\: ....\\\\ &\text{a}.\quad \displaystyle \frac{25}{24}\qquad\qquad\qquad\text{d}.\quad \displaystyle \frac{1}{4}\\ &\text{b}.\quad \displaystyle \frac{24}{25}\qquad\qquad\quad\quad\text{e}.\quad \displaystyle \frac{1}{12}\\ &\text{c}.\quad \displaystyle \frac{7}{24}\\\\ &\textbf{Jawab:  c}\\ &\begin{aligned}&\textrm{Perhatikan bahwa deret jumlah semua sukunya}:\\ &\displaystyle \frac{1}{5}+\frac{2}{5^{\displaystyle 2}}+\frac{1}{5^{\displaystyle 3}}+\frac{2}{5^{\displaystyle 4}}+\frac{1}{5^{\displaystyle 5}}+\frac{2}{5^{\displaystyle 6}}\:+\:...\\ &\textrm{Selanjutnya deret kita partisi menjadi dua bagian, yaitu}:\\ &\textbf{Deret pertama untuk suku ganjil}\\ &S_{\displaystyle 1}=\displaystyle \frac{1}{5}+\frac{1}{5^{\displaystyle 3}}+\frac{1}{5^{\displaystyle 5}}\:+\:...\\ &\begin{cases} \textrm{suku ke}-1&= U_{\displaystyle 1}=a=\displaystyle \frac{1}{5} \\\textrm{rasio}&=r=\displaystyle \frac{1}{25}\\ S_{\displaystyle \infty }&=\displaystyle \frac{a}{1-r}=\frac{\displaystyle \frac{1}{5}}{1-\displaystyle \frac{1}{25}}=\displaystyle \frac{5}{24} \end{cases}\\ &\textbf{Deret kedua untuk suku genap}\\ &S_{\displaystyle 2}=\frac{2}{5^{\displaystyle 2}}+\frac{2}{5^{\displaystyle 4}}+\frac{2}{5^{\displaystyle 6}}\:+\:...\\ &\begin{cases} \textrm{suku ke}-1&= U_{\displaystyle 1}=a=\displaystyle \frac{2}{5^{\displaystyle 2}}=\frac{2}{25} \\\textrm{rasio}&=r=\displaystyle \frac{1}{25}\\ S_{\displaystyle \infty }&=\displaystyle \frac{a}{1-r}=\frac{\displaystyle \frac{2}{25}}{1-\displaystyle \frac{1}{25}}=\displaystyle \frac{2}{24} \end{cases}\\ &\textbf{Jumlah totalnya adalah}:\\ &S_{\textrm{total}}=S_{\displaystyle 1}+S_{\displaystyle 2}=\displaystyle \frac{5}{24}+\frac{2}{24}=\displaystyle \frac{7}{24}\end{aligned}\end{aligned}$.

$\begin{aligned}6.\quad&(\textbf{KSM Matematika MA Tk. Kab/kota Th.2013})\\ &\textrm{nilai}\quad n\quad \textrm{terkecil yang memenuhi}\:\: \frac{1}{2^{\displaystyle n}}\lt 0,001\:\: \textrm{adalah}\: ....\\ &\text{a}.\quad \displaystyle 9\qquad\qquad\qquad\text{d}.\quad \displaystyle 522\\ &\text{b}.\quad \displaystyle 10\:\:\quad\quad\quad\quad\quad\text{e}.\quad \displaystyle 501\\ &\text{c}.\quad \displaystyle 11\\\\ &\textbf{Jawab:  b}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}:\\ &\frac{1}{2^{\displaystyle n}}\lt 0,001\Leftrightarrow \displaystyle \frac{1}{2^{\displaystyle n}}\lt \displaystyle \frac{1}{1000}\Leftrightarrow 2^{\displaystyle n}\gt 1000\\ &\textrm{Selanjutnya cukup kita uji untuk}\\&\begin{cases} n=9&\Rightarrow 2^{\displaystyle 9}=512\ngtr 1000 \\n=10&\Rightarrow 2^{\displaystyle 10}=1024\gt  1000\\ n=11&\Rightarrow 2^{\displaystyle 11}=2048\gt  1000\: (\textrm{bukan yang terkecil}) \end{cases}\\ &\textbf{Jadi,}\:\: n\:\: \textrm{terkecilnya adalah}=10\end{aligned}\end{aligned}$.