Misalkan \( 1+\frac{1}{p}+\frac{1}{p^3}+\cdots = 2021 \) dan \( \frac{1}{q}+\frac{1}{q^2}+\frac{1}{q^3}+\cdots = 2019 \). Nilai dari \( \frac{q}{p} = \cdots \)
- \( \frac{2018 \times 2020}{ 2021 \times 2019 } \)
- \( \frac{2019 \times 2020}{ 2018 \times 2021 } \)
- \( \frac{2020^2 - 1}{ 2020^2 } \)
- \( \frac{2020^2}{ 2020^2-1 } \)
- \( \frac{2020^2}{ 2020^2+1 } \)
Pembahasan:
Dari deret \( 1+\frac{1}{p}+\frac{1}{p^3}+\cdots = 2021 \), kita peroleh \(a = 1\) dan \(r = \frac{1}{p}\) sehingga dapat dituliskan:
\begin{aligned} S_\infty = \frac{a}{1-r} &\Leftrightarrow 2021 = \frac{1}{1-\frac{1}{p}} \\[8pt] 2021 = \frac{1}{\frac{p-1}{p}} &\Leftrightarrow 2021 = \frac{p}{p-1} \\[8pt] 2021p-2021 &= p \\[8pt] 2020p &= 2021 \\[8pt] p &= \frac{2021}{2020} \end{aligned}
Dari deret \( \frac{1}{q}+\frac{1}{q^2}+\frac{1}{q^3}+\cdots = 2019 \), kita peroleh \( a = \frac{1}{q} \) dan \( r = \frac{1}{q} \) sehingga dapat dituliskan:
\begin{aligned} S_\infty = \frac{a}{1-r} &\Leftrightarrow 2019 = \frac{\frac{1}{q}}{1-\frac{1}{q}} \\[8pt] 2019 = \frac{\frac{1}{q}}{\frac{q-1}{q}} &\Leftrightarrow 2019 = \frac{1}{q-1} \\[8pt] 2019q-2019 &= 1 \\[8pt] 2019q &= 2020 \\[8pt] q &= \frac{2020}{2019} \end{aligned}
Dari nilai \(p\) dan \(q\) di atas, maka diperoleh:
\begin{aligned} \frac{q}{p} &= \frac{ \frac{2020}{2019} }{ \frac{2021}{2020} } = \frac{2020}{2019} \times \frac{2020}{2021} \\[8pt] &= \frac{2020^2}{ (2020+1) \times (2020-1) } \\[8pt] &= \frac{2020^2}{2020^2-1} \end{aligned}
Jawaban D.