Pertanyaan :
Misalkan suatu deret geometri, S = 2 + [tex]\sqrt{\frac{4c^2-24}{5c}}+\frac{2c^2-12}{5c}+......[/tex] bila S ∈ R maka nilai c yang memenuhi adalah ?
a. c < -1 atau √6 < c < 6
b. -1 < c < √6
c. -√6 < c < - 1 atau 0 < c < 6
d. 0 < c < 6
e. √6 < c < -1 atau √6 < o < 6
Misalkan suatu deret geometri
[tex]\begin{aligned}S=2+\sqrt{\frac{4c^2-24}{5c}}+\frac{2c^2-12}{5c}+{\dots}\end{aligned}[/tex]
Apabila [tex]S\in\mathbb{R}[/tex], maka nilai [tex]c[/tex] yang memenuhi adalah:
[tex]\boxed{\large\text{$\begin{aligned}\bf{-}\sqrt{6} < c < -1\ \:{\sf atau}\ \:\sqrt{6} < c < 6\end{aligned}$}}[/tex]
Pembahasan
Deret geometri tersebut dapat dinyatakan juga dengan:
[tex]\begin{aligned}S&=2+\sqrt{\frac{4\left(c^2-6\right)}{5c}}+\frac{2\left(c^2-6\right)}{5c}+{\dots}\\\Rightarrow S&=2+2\sqrt{\frac{c^2-6}{5c}}+2\left(\sqrt{\frac{c^2-6}{5c}}\right)^2+{\dots}\\\end{aligned}[/tex]
Sehingga, jelas bahwa pada deret geometri tersebut,
[tex]\begin{aligned}a=U_1=2\,,\ \ r=\sqrt{\frac{c^2-6}{5c}}\end{aligned}[/tex]
Agar [tex]S\in\mathbb{R}[/tex], dan terbatas (syarat implisit), syaratnya adalah [tex]S[/tex] konvergen, sehingga [tex]r[/tex] harus berada dalam rentang [tex]-1 < r < 1[/tex] dengan [tex]r\in\mathbb{R}[/tex]. Perlu diperhatikan pula bahwa agar [tex]S[/tex] merupakan sebuah deret, deret [tex]S[/tex] memuat suku-suku tak-nol, sehingga [tex]r \ne 0[/tex].
Agar [tex]r\in\mathbb{R}[/tex] dan [tex]r \ne 0[/tex]:
[tex]\begin{aligned}&\frac{c^2-6}{5c}\ > \ 0\quad\Big\}\textsf{ faktorkan}\\&\Rightarrow \frac{\left(c+\sqrt{6}\right)\left(c-\sqrt{6}\right)}{5c}\ > \ 0\,,\ c\ne 0\\&\Rightarrow \frac{\left(c+\sqrt{6}\right)\left(c-\sqrt{6}\right)}{5c}\cdot5\ > \ 0\cdot5\,,\ c\ne 0\\&\Rightarrow \frac{\left(c+\sqrt{6}\right)\left(c-\sqrt{6}\right)}{c}\ > \ 0\,,\ c\ne 0\end{aligned}[/tex]
[tex]\begin{aligned}&\Rightarrow\left[\ \begin{aligned}\bullet\ &c < -\sqrt{6}:\\&\Rightarrow\frac{(-)(-)}{(-)} < 0\\&\Rightarrow \textsf{tidak memenuhi}\\\bullet\ &{-}\sqrt{6} < c < 0:\\&\Rightarrow\frac{(+)(-)}{(-)} > 0\\&\Rightarrow \textsf{memenuhi}\\\bullet\ &0 < c < \sqrt{6}:\\&\Rightarrow\frac{(+)(-)}{(+)} < 0\\&\Rightarrow \textsf{tidak memenuhi}\\\bullet\ &c > \sqrt{6}:\\&\Rightarrow\frac{(+)(+)}{(+)} > 0\\&\Rightarrow \textsf{memenuhi}\\\end{aligned}\right.\end{aligned}[/tex]
[tex]\begin{aligned}&\therefore\ \sf Rentang\ yang\ memenuhi:\\&\qquad{-}\sqrt{6} < c < 0\ {\sf atau}\ c > \sqrt{6}\\&\qquad\Rightarrow\ \big({-}\sqrt{6},\ 0\big)\cup\big(\sqrt{6},\ \infty\big)\quad...(i)\end{aligned}[/tex]
Kemudian, [tex]-1 < r < 1[/tex] berarti:
[tex]\begin{aligned}&{-}1 < \sqrt{\frac{c^2-6}{5c}} < 1\\&\Rightarrow -1 < \sqrt{\frac{c^2-6}{5c}}\ \;{\sf dan}\ \;\sqrt{\frac{c^2-6}{5c}} < 1\end{aligned}[/tex]
Karena [tex]r=\sqrt{\dfrac{c^2-6}{5c}} > 0[/tex], maka [tex]-1 < \sqrt{\dfrac{c^2-6}{5c}}[/tex] benar dan berlaku untuk semua [tex]c \in \mathbb{R}[/tex].
Lalu,
[tex]\begin{aligned}&\sqrt{\frac{c^2-6}{5c}} < 1\\&\Rightarrow\frac{c^2-6}{5c} < 1\ \ {\sf dan}\ \ \frac{c^2-6}{5c}\ \ge\ 0\\&\Rightarrow\frac{c^2-6}{5c} - 1 < 0\\&\Rightarrow\frac{c^2-6-5c}{5c} < 0 \\&\Rightarrow\frac{c^2-6-5c}{5c}\cdot5 < 0\cdot5 \\&\Rightarrow\frac{c^2-6-5c}{c} < 0 \\&\Rightarrow\frac{(c+1)(c-6)}{c} < 0\end{aligned}[/tex]
[tex]\begin{aligned}&\Rightarrow\left[\ \begin{aligned}&c < -1:\\&\Rightarrow \frac{(-)(-)}{(-)} < 0\\&\Rightarrow \textsf{memenuhi}\\&{-}1 \le c < 0:\\&\Rightarrow \frac{(+\lor0)(-)}{(-)} \ge 0\\&\Rightarrow \textsf{tidak memenuhi}\\&0 < c < 6:\\&\Rightarrow \frac{(+)(-)}{(+)} < 0\\&\Rightarrow \textsf{memenuhi}\\&c \ge 6:\\&\Rightarrow \frac{(+)(+\lor0)}{(+)} \ge 0\\&\Rightarrow \textsf{tidak memenuhi}\\\end{aligned}\right.\end{aligned}[/tex]
[tex]\begin{aligned}&\therefore\ {\sf Rentang\ yang\ memenuhi}\ \frac{c^2-6}{5c} < 1:\\&\qquad c < -1\ {\sf atau}\ 0 < c < 6\\&\qquad\Rightarrow\ \big({-}\infty,\ -1\big)\cup\big(0,\ 6\big)\quad...(ii)\end{aligned}[/tex]
Sehingga, rentang nilai [tex]c[/tex] untuk deret geometri
[tex]\begin{aligned}S=2+\sqrt{\frac{4c^2-24}{5c}}+\frac{2c^2-12}{5c}+{\dots}\end{aligned}[/tex]
bila [tex]S\in\mathbb{R}[/tex] adalah:
[tex]\begin{aligned}(i)\cap(ii)&=\big({-}\sqrt{6},\ 0\big)\cup\big(\sqrt{6},\ \infty\big)\\&\qquad\qquad\quad\bigcap\\&\quad\ \ \big({-}\infty,\ -1\big)\cup\big(0,\ 6\big)\\&=\big({-}\sqrt{6},\ -1\big)\:\cup\:\big(\sqrt{6},\ 6\big)\\\end{aligned}[/tex]
atau dapat dinyatakan juga dengan:
[tex]\boxed{\large\text{$\begin{aligned}\bf{-}\sqrt{6} < c < -1\ \:{\sf atau}\ \:\sqrt{6} < c < 6\end{aligned}$}}[/tex]
[tex]\blacksquare[/tex]
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