Lanjutan (Barisan & Deret)

$\begin{array}{l}\\ 5.& \mathbf{\text{(Lomba Matematika Nasional}}\\ & \mathbf{\text{HIMATIKA UGM 2006)}}\\ & \text{Jika bilangan}\\ & A={\displaystyle \frac{1}{1+1}+\frac{1}{1+2}+\frac{1}{1+3}+\cdots +\frac{1}{1+100}}\\ & B={\displaystyle \frac{1}{1+1}+\frac{1}{1+{\displaystyle \frac{1}{2}}}+\frac{1}{1+{\displaystyle \frac{1}{3}}}+\cdots +\frac{1}{1+{\displaystyle \frac{1}{100}}}}\\ & \text{maka}A+B\text{sama dengan}....\\ & \text{A}.202\text{D}.100\\ & \text{B}.200\text{C}.101\text{E}.99\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{l}\\ \begin{array}{rl}A& ={\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{101}}\\ B& ={\displaystyle \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots +\frac{100}{101}}\end{array}& \\ & +\\ \\ A+B=\underset{100}{\underbrace{1+1+1+\cdots +1}}& =100\end{array}\end{array}$.

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

 $\begin{array}{l}\\ 7.& \text{Misalkan}\\ & a_n={\displaystyle \frac{n(n+1)}{2},\text{tentukanlah jumlah}}\\ & \text{dari}{\displaystyle \frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{2023}}}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}{a}_n& ={\displaystyle \frac{n(n+1)}{2},\text{maka}}\\ {\displaystyle \frac{1}{a_n}}& ={\displaystyle \frac{2}{n(n+1)}}\\ & =2\left({\displaystyle \frac{1}{n}-\frac{1}{n+1}}\right)\\ & \text{lihat pembahasan no.3 di atas}\\ & \end{array}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{2023}}}\\ & =2((1-{\displaystyle \frac{1}{2}})+\left({\displaystyle \frac{1}{2}-\frac{1}{3}}\right)+\cdots +\left({\displaystyle \frac{1}{2022}-\frac{1}{2023}}\right))\\ & =2(1-{\displaystyle \frac{1}{2023}})\\ & =2\left({\displaystyle \frac{2022}{2023}}\right)={\displaystyle \frac{4044}{2023}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Misalkan}n\text{adalah bilangan asli}\\ & \text{dan}\left\{a_n\right\}\text{adalah barisan bilangan real}\\ & \text{dengan}{a}_n={\displaystyle \frac{2^n}{2^{2n+1}-2^{n+1}-2^n+1}}\\ \\ & \text{Tunjukkan bahwa untuk setiap bilangan}\\ & \text{asli}n,\text{berlaku}{a}_1+a_2+\cdots +a_n<1\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}{a}_n& ={\displaystyle \frac{2^n}{2^{2n+1}-2^{n+1}-2^n+1}}\\ & ={\displaystyle \frac{2^n}{(2^{n+1}-1)(2^n-1)}}\\ & =\left({\displaystyle \frac{1}{2^{n+1}-1}-\frac{1}{2^n-1}}\right)\\ a_1& ={\displaystyle \frac{1}{2^1-1}-\frac{1}{2^2-1}=1-{\displaystyle \frac{1}{3}}}\\ a_2& ={\displaystyle \frac{1}{2^2-1}-\frac{1}{2^3-1}=\frac{1}{3}-\frac{1}{7}}\\ a_3& ={\displaystyle \frac{1}{2^3-1}-\frac{1}{2^4-1}=\frac{1}{7}-\frac{1}{15}}\\ & ⋮⋮\\ a_n& ={\displaystyle \frac{1}{2^n-1}-\frac{1}{2^{n+1}-1}+}\\ a_1& +a_2+a_3+\cdots +a_n\\ & =1-{\displaystyle \frac{1}{2^{n+1}-1}<1◼}\end{array}\end{array}$.