等比級數2008-1013-1508

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$S_{n}=\underbrace{2}_{1}+\underbrace{22}_{2}+\underbrace{222}_{3}+\cdots +\underbrace{2\cdots 2}_{n}$
$\Rightarrow S_{n}=\frac{2}{9}\left( \underbrace{9}_{1}+\underbrace{99}_{2}+\underbrace{999}_{3}+\cdots +\underbrace{9\cdots 9}_{n} \right)$
$=\frac{2}{9}\left[ \left( 10-1 \right)+\left( 10^{2}-1 \right)+\cdots +\left( 10^{n}-1 \right) \right]$
$=\frac{2}{9}\left[ \left( 10+10^{2}+\cdots +10^{n} \right)-n \right]$
$=\frac{2}{9}\left[ \frac{10\left( 10^{n}-1 \right)}{10-1}-n \right]=\frac{2}{9}\times \frac{\left( 10^{n+1}-10 \right)}{9}-\frac{2}{9}n$
$=\frac{2}{81}\times 10^{n+1}-\frac{20}{81}-\frac{2}{9}n\#$

$S_{n}=\underbrace{0.2}_{1}+\underbrace{0.22}_{2}+\underbrace{0.222}_{3}+\cdots +\underbrace{0.2\cdots 2}_{n}$
$\Rightarrow S_{n}=\frac{2}{9}\left( \underbrace{0.9}_{1}+\underbrace{0.99}_{2}+\underbrace{0.999}_{3}+\cdots +\underbrace{0.9\cdots 9}_{n} \right)$
$=\frac{2}{9}\left[ \left( 1-0.1 \right)+\left( 1-0.1^{2} \right)+\left( 1-0.1^{3} \right)+\cdots +\left( 1-0.1^{n} \right) \right]$
$=\frac{2}{9}\left[ n-\left( 0.1+0.1^{2}+\cdots +0.1^{n} \right) \right]=\frac{2}{9}\left\{ n-\left[ \frac{0.1\left( 1-0.1^{n} \right)}{1-0.1} \right] \right\}$
$=\frac{2}{9}n-\frac{2}{9}\left[ \frac{0.1\left( 1-0.1^{n} \right)}{0.9}\times \frac{10}{10} \right]=\frac{2}{9}n-\frac{2}{9}\left( \frac{1-0.1^{n}}{9} \right)$
$=\frac{2}{9}n-\frac{2}{81}+\frac{2}{81}\times \left( 0.1^{n} \right)\#$