Lanjutan (Barisan & Deret)

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

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

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

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