數學：等比級數n項和與2n項和及3n項和為等比數列之證明

出自高材生

已知： $\left\{ \begin{align} & \left\{ a_{n} \right\}\to \text{a geometric sequence} \\ & S_{n}=\sum\limits_{k=1}^{n}{a_{k}},S_{2n}=\sum\limits_{k=1}^{2n}{a_{k}},S_{3n}=\sum\limits_{k=1}^{3n}{a_{k}} \\ \end{align} \right.$

求證： $\Rightarrow S_{n},S_{2n}-S_{n},S_{3n}-S_{2n}\to \text{a geometric sequence}$

$\left\{ \begin{align} & S_{n}=\frac{a_{1}\times \left( 1-r^{n} \right)}{1-r} \\ & S_{2n}=\frac{a_{1}\times \left( 1-r^{2n} \right)}{1-r} \\ & S_{3n}=\frac{a_{1}\times \left( 1-r^{3n} \right)}{1-r} \\ \end{align} \right.$

$S_{2n}-S_{n}=\frac{a_{1}\times \left( 1-r^{2n} \right)}{1-r}-\frac{a_{1}\times \left( 1-r^{n} \right)}{1-r}=\frac{a_{1}\times \left( r^{n}-r^{2n} \right)}{1-r}=\frac{a_{1}\times r^{n}\times \left( 1-r^{n} \right)}{1-r}$

$\frac{S_{2n}-S_{n}}{S_{n}}=\frac{\frac{a_{1}\times r^{n}\times \left( 1-r^{n} \right)}{1-r}}{\frac{a_{1}\times \left( 1-r^{n} \right)}{1-r}}={\color{Red}r^{n}}$

$S_{3n}-S_{2n}=\frac{a_{1}\times \left( 1-r^{3n} \right)}{1-r}-\frac{a_{1}\times \left( 1-r^{2n} \right)}{1-r}=\frac{a_{1}\times \left( r^{2n}-r^{3n} \right)}{1-r}=\frac{a_{1}\times r^{2n}\left( 1-r^{n} \right)}{1-r}$

$\frac{S_{3n}-S_{2n}}{S_{2n}-S_{n}}=\frac{\frac{a_{1}\times r^{2n}\left( 1-r^{n} \right)}{1-r}}{\frac{a_{1}\times r^{n}\times \left( 1-r^{n} \right)}{1-r}}={\color{Red}r^{n}}=\frac{S_{2n}-S_{n}}{S_{n}}$
$\Rightarrow S_{n},S_{2n}-S_{n},S_{3n}-S_{2n}\to \text{a geometric sequence}\#$

取自"http://www.come2pass.com/wiki/index.php/%E6%95%B8%E5%AD%B8%EF%BC%9A%E7%AD%89%E6%AF%94%E7%B4%9A%E6%95%B8n%E9%A0%85%E5%92%8C%E8%88%872n%E9%A0%85%E5%92%8C%E5%8F%8A3n%E9%A0%85%E5%92%8C%E7%82%BA%E7%AD%89%E6%AF%94%E6%95%B8%E5%88%97%E4%B9%8B%E8%AD%89%E6%98%8E"

數學：等比級數n項和與2n項和及3n項和為等比數列之證明

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