Lanjutan 3 (Barisan & Deret)

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

$\begin{array}{l}\\ 12.& \text{Diketahui}\\ & x={\displaystyle \frac{1+p+p^2+p^3+\cdots +p^{n-1}}{1+p+p^2+p^3+\cdots +p^{n-2}+p^{n-1}+p^n}}\\ & y={\displaystyle \frac{1+q+q^2+q^3+\cdots +q^{n-1}}{1+q+q^2+q^3+\cdots +q^{n-2}+q^{n-1}+q^n}}\\ \\ & \text{dan}p>q>0\\ \\ & \text{Tunjukkan bahwa}x<y\\ \\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa}:p>q>0\\ & \text{sehingga}\\ & {\displaystyle \frac{1}{p}<\frac{1}{q},{\displaystyle \frac{1}{p^2}<\frac{1}{q^2},\cdots ,{\displaystyle \frac{1}{p^n}<\frac{1}{q^n}}}}\\ & \text{Jika bentuk di atas dijumlahkan, maka}\\ & {\displaystyle \frac{1}{p}+\frac{1}{p^2}+\cdots +\frac{1}{p^n}<\frac{1}{q}+\frac{1}{q^2}+\cdots +\frac{1}{q^n}}\\ & \iff {\displaystyle \frac{p^{n-1}+\cdots +p^2+p+1}{p^n}<{\displaystyle \frac{q^{n-1}+\cdots +q^2+q+1}{q^n}}}\\ & \iff {\displaystyle \frac{p^n}{1+p+p^2+\cdots +p^{n-1}}>{\displaystyle \frac{q^n}{1+q+q^2+\cdots +q^{n-1}}}}\\ & \iff {\displaystyle \frac{p^n}{1+p+p^2+\cdots +p^{n-1}}+1>{\displaystyle \frac{q^n}{1+q+q^2+\cdots +q^{n-1}}+1}}\\ & \iff {\displaystyle \frac{1+p+p^2+\cdots +p^{n-1}+p^n}{1+p+p^2+\cdots +p^{n-1}}>{\displaystyle \frac{1+q+q^2+\cdots +q^{n-1}+q^n}{1+q+q^2+\cdots +q^{n-1}}}}\\ & \iff {\displaystyle \frac{1+p+p^2+\cdots +p^{n-1}}{1+p+p^2+\cdots +p^{n-1}+p^n}<{\displaystyle \frac{1+q+q^2+\cdots +q^{n-1}}{1+q+q^2+\cdots +q^{n-1}+q^n}}}\\ & \iff x<y◼\end{array}\end{array}$.

DAFTAR PUSTAKA

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