두 등비급수의 곱 조건

등비수열 $\left\{ a _{n} \right\}$의 첫째항을 $a _{1}$, 공비를 $r _{1}$, 등비수열 $\left\{ b _{n} \right\}$의 첫째항을 $b _{1}$, 공비를 $r _{2}$라 하면 두 급수 $\displaystyle \sum\limits _{n = 1} ^{\infty} a _{n}$, $\displaystyle \sum\limits _{n = 1} ^{\infty} b _{n}$이 각각 수렴하므로 $- 1 < r _{1} < 1$, $- 1 < r _{2} < 1$ $\therefore$ $\displaystyle \sum\limits _{n = 1} ^{\infty} a _{n}$$\displaystyle = \frac{a _{1}}{1 - r _{1}}$, $\displaystyle \sum\limits _{n = 1} ^{\infty} b _{n}$$\displaystyle = \frac{b _{1}}{1 - r _{2}}$ 이때 수열 $\left\{ a _{n} b _{n} \right\}$은 첫째항이 $a _{1} b _{1}$, 공비가 $r _{1} r _{2}$인 등비수열이고 $- 1 < r _{1} r _{2} < 1$이므로 $\displaystyle \sum\limits _{n = 1} ^{\infty} a _{n} b _{n}$$\displaystyle = \frac{a _{1} b _{1}}{1 - r _{1} r _{2}}$ 즉 $\displaystyle \sum\limits _{n = 1} ^{\infty} a _{n} b _{n}$$\displaystyle = \sum\limits _{n = 1} ^{\infty} a _{n} \sum\limits _{n = 1} ^{\infty} b _{n}$에서 $\displaystyle \frac{a _{1} b _{1}}{1 - r _{1} r _{2}}$$\displaystyle = \frac{a _{1}}{1 - r _{1}} \times \frac{b _{1}}{1 - r _{2}}$ $\therefore$ $r _{1} + r _{2} = 2 r _{1} r _{2}$ $\cdots$$\cdots$ ㉠ 또 수열 $\left\{ \left| a _{2n} \right| \right\}$은 첫째항이 $\left| a _{1} r _{1} \right|$, 공비가 $r _{1} \mathit{^{2}}$인 등비수열이고, 수열 $\left\{ \left| a _{3n} \right| \right\}$은 첫째항이 $\left| a _{1} r _{1\mathit{^{2}}} \right|$, 공비가 $\left| r _{1} \mathit{^{3}} \right|$인 등비수열이며 $0 \leq r _{1} \mathit{^{2}} < 1$, $0 \leq \left| r _{1} \mathit{^{3}} \right| < 1$이므로 $\displaystyle \sum\limits _{n = 1} ^{\infty} \left| a _{2n} \right|$$\displaystyle = \frac{\left| a _{1} r _{1} \right|}{1 - r _{1\mathit{^{2}}}}$, $\displaystyle \sum\limits _{n = 1} ^{\infty} \left| a _{3n} \right|$$\displaystyle = \frac{\left| a _{1} r _{1} ^{2} \right|}{1 - \left| r _{1} \mathit{^{3}} \right|}$ 따라서 $\displaystyle 3 \times \sum\limits _{n = 1} ^{\infty} \left| a _{2n} \right| = 7 \times \sum\limits _{n = 1} ^{\infty} \left| a _{3n} \right|$에서 $\displaystyle \frac{3 \left| a _{1} r _{1} \right|}{1 - r _{1\mathit{^{2}}}} = \frac{7 \left| a _{1} r _{1} ^{2} \right|}{1 - \left| r _{1} \mathit{^{3}} \right|}$ (ⅰ) $r _{1} > 0$일 때 $\displaystyle \frac{3}{1 - r _{1} ^{2}} = \frac{7 r _{1}}{1 - r _{1} ^{3}}$에서 $3 - 3 r _{1} ^{3} = 7 r _{1} - 7 r _{1} ^{3}$, $\left( 2 r _{1} + 3 \right) \left( 2 r _{1} - 1 \right) \left( r _{1} - 1 \right) = 0$ $\therefore$ $\displaystyle r _{1} = \frac{1}{2}$ 즉 ㉠에서 $\displaystyle \frac{1}{2} + r _{2} = 2 \times \frac{1}{2} \times r _{2}$이므로 이를 만족시키는 $r _{2}$는 존재하지 않는다. (ⅱ) $r _{1} < 0$일 때 $\displaystyle \frac{3}{1 - r _{1} ^{2}} = \frac{- 7 r _{1}}{1 + r _{1} ^{3}}$에서 $3 + 3 r _{1} ^{3} = - 7 r _{1} + 7 r _{1} ^{3}$, $\left( 2 r _{1} - 3 \right) \left( 2 r _{1} + 1 \right) \left( r _{1} + 1 \right) = 0$ $\therefore$ $\displaystyle r _{1} = - \frac{1}{2}$ 즉 ㉠에서 $\displaystyle \left( - \frac{1}{2} \right) + r _{2} = 2 \times \left( - \frac{1}{2} \right) r _{2}$이므로 $\displaystyle r _{2} = \frac{1}{4}$ (ⅰ), (ⅱ)에 의하여 $\displaystyle \sum\limits _{n = 1} ^{\infty} \frac{b _{2n-1} + b _{3n+1}}{b _{n}}$ $\displaystyle = \sum\limits _{n = 1} ^{\infty} \frac{b _{1} \left\{ \left( \frac{1}{4} \right) ^{2} \right\} ^{n-1} + b _{1} \left\{ \left( \frac{1}{4} \right) ^{3} \right\} ^{n}}{b _{1} \left( \frac{1}{4} \right) ^{n-1}}$ $\displaystyle = \sum\limits _{n = 1} ^{\infty} \left\{ \left( \frac{1}{4} \right) ^{n-1} + \left( \frac{1}{4} \right) ^{2n+1} \right\}$ $\displaystyle = \frac{1}{1 - \frac{1}{4}} + \frac{\frac{1}{64}}{1 - \frac{1}{16}}$ $\displaystyle = \frac{4}{3} + \frac{1}{60}$ $\displaystyle = \frac{27}{20}$ 따라서 $\displaystyle S = \frac{27}{20}$이므로 $120 S$$\displaystyle = 120 \times \frac{27}{20}$$= 162$