Selingan (Barisan & Deret)

Lanjutan contoh soal dan pembahasannya terkait barisan dan deret

$\begin{array}{l}\\ 1.& \text{Hasil kali bilangan bentuk berikut}\\ & (1-{\displaystyle \frac{1}{4}})(1-{\displaystyle \frac{1}{5}})(1-{\displaystyle \frac{1}{6}})\cdots (1-{\displaystyle \frac{1}{100}})\\ & \text{adalah}....\\ & \text{A}.{\displaystyle \frac{1}{100}\text{D}.{\displaystyle \frac{4}{100}}}\\ \\ & \text{B}.{\displaystyle \frac{2}{100}\text{C}.{\displaystyle \frac{3}{100}\text{E}.{\displaystyle \frac{5}{100}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{rl}& (1-{\displaystyle \frac{1}{4}})(1-{\displaystyle \frac{1}{5}})(1-{\displaystyle \frac{1}{6}})\cdots (1-{\displaystyle \frac{1}{100}})\\ & =\left({\displaystyle \frac{3}{4}}\right)\left({\displaystyle \frac{4}{5}}\right)\left({\displaystyle \frac{5}{6}}\right)\cdots \left({\displaystyle \frac{99}{100}}\right)\\ & =\left({\displaystyle \frac{3}{\text{⧸}4}}\right)\left({\displaystyle \frac{\text{⧸}4}{\text{⧸}5}}\right)\left({\displaystyle \frac{\text{⧸}5}{\text{⧸}6}}\right)\cdots \left({\displaystyle \frac{\text{⧸}99}{100}}\right)\\ & ={\displaystyle \frac{3}{100}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Hasil kali bilangan bentuk berikut}\\ & (1-{\displaystyle \frac{2}{3}})(1-{\displaystyle \frac{2}{5}})(1-{\displaystyle \frac{2}{7}})\cdots (1-{\displaystyle \frac{2}{2023}})\\ & \text{adalah}....\\ & \text{A}.{\displaystyle \frac{1}{2023}\text{D}.{\displaystyle \frac{4}{2023}}}\\ \\ & \text{B}.{\displaystyle \frac{2}{2023}\text{C}.{\displaystyle \frac{3}{2023}\text{E}.{\displaystyle \frac{5}{2023}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dengan cara pembahasan pada no.1 di atas, maka}\\ & \begin{array}{rl}& (1-{\displaystyle \frac{2}{3}})(1-{\displaystyle \frac{2}{5}})(1-{\displaystyle \frac{2}{7}})\cdots (1-{\displaystyle \frac{2}{2023}})\\ & =\left({\displaystyle \frac{1}{3}}\right)\left({\displaystyle \frac{3}{5}}\right)\left({\displaystyle \frac{5}{7}}\right)\cdots \left({\displaystyle \frac{2021}{2023}}\right)\\ & =\left({\displaystyle \frac{1}{\text{⧸}3}}\right)\left({\displaystyle \frac{\text{⧸}3}{\text{⧸}5}}\right)\left({\displaystyle \frac{\text{⧸}5}{\text{⧸}7}}\right)\cdots \left({\displaystyle \frac{\text{⧸}2021}{2023}}\right)\\ & ={\displaystyle \frac{1}{2023}}\end{array}\end{array}$.

 $\begin{array}{l}\\ 3.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{2022\times 2023}}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& =(1-{\displaystyle \frac{1}{2}})+\left({\displaystyle \frac{1}{2}-\frac{1}{3}}\right)+\left({\displaystyle \frac{1}{3}-\frac{1}{4}}\right)+\\ & +\cdots +\left({\displaystyle \frac{1}{2022}-\frac{1}{2023}}\right)\\ & =1-{\displaystyle \frac{1}{2023}}\\ & ={\displaystyle \frac{2022}{2023}}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 4.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{2023}-\sqrt{2022}}}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{1+\sqrt{2}}}& =\frac{1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}\\ & ={\displaystyle \frac{1-\sqrt{2}}{1^2-(\sqrt{2}{)}^2}=\frac{1-\sqrt{2}}{1-2}}\\ & ={\displaystyle \frac{1-\sqrt{2}}{-1}=\sqrt{2}-1}\\ {\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}}}& =\frac{1}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\\ & ={\displaystyle \frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}{)}^2-(\sqrt{3}{)}^2}=\frac{\sqrt{2}-\sqrt{3}}{2-3}}\\ & ={\displaystyle \frac{\sqrt{2}-\sqrt{3}}{-1}=\sqrt{3}-\sqrt{2}}\\ {\displaystyle \frac{1}{\sqrt{3}+\sqrt{4}}}& =\frac{1}{\sqrt{3}+\sqrt{4}}\times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\\ & ={\displaystyle \frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3}{)}^2-(\sqrt{4}{)}^2}=\frac{\sqrt{3}-\sqrt{4}}{3-4}}\\ & ={\displaystyle \frac{\sqrt{3}-\sqrt{4}}{-1}=\sqrt{4}-\sqrt{3}}\\ & ⋮\end{array}\\ & \text{Sehingga}\\ & \begin{array}{rl}& =\text{⧸}\sqrt{2}-1+\text{⧸}\sqrt{3}-\text{⧸}\sqrt{2}+\text{⧸}\sqrt{4}-\text{⧸}\sqrt{3}+\cdots +\sqrt{2023}-\text{⧸}\sqrt{2022}\\ & =\sqrt{2023}-1\end{array}\end{array}$