数列

$\frac{23}{111}$を$0.a_1a_2a_3a_4\cdots$のように小数で表す．
すなわち，小数第$k$位の数を$a_k$とする．このとき、$S=\sum_{k=1}^{n}\left(\frac{a_k}{3}\right)^k$を求めよ．


$\frac{23}{111}= 0.207207207207207\cdots$なので，
$a_1=2,\ a_2=0,\ a_3=7,\ $
$a_4=2,\ a_5=0,\ a_6=7,\ $
$a_7=2,\ a_8=0,\ a_9=7,\ $・・・
よって，
$S=(\frac{a_1}{3})^{1}+(\frac{a_2}{3})^{2}+(\frac{a_3}{3})^{3}+(\frac{a_4}{3})^{4}+\cdots$
$=(\frac{2}{3})^{1}+(\frac{0}{3})^{2}+(\frac{7}{3})^{3}+(\frac{2}{3})^{4}+(\frac{0}{3})^{5}+(\frac{7}{3})^{6}+(\frac{2}{3})^{7}+(\frac{0}{3})^{8}+(\frac{7}{3})^{9}+\cdots$

nが3の倍数 $n=3m$ のとき．
$S=\left((\frac{2}{3})^1+(\frac{0}{3})^2+(\frac{7}{3})^3\right)+\left((\frac{2}{3})^4+(\frac{0}{3})^5+(\frac{7}{3})^6\right)+\\ \cdots+\left((\frac{2}{3})^{n-2}+(\frac{0}{3})^{n-1}+(\frac{7}{3})^{n}\right)$
$=\left((\frac{2}{3})^1+(\frac{2}{3})^4+(\frac{2}{3})^7+\cdots+(\frac{2}{3})^{3m-2}\right)\\ +\left((\frac{0}{3})^2+(\frac{0}{3})^5+(\frac{0}{3})^8+\cdots+(\frac{0}{3})^{3m-1}\right)\\ +\left((\frac{7}{3})^3+(\frac{7}{3})^6+(\frac{7}{3})^9+\cdots+(\frac{7}{3})^{3m}\right)$
＝（初項$\frac{2}{3}$，公比$\frac{1}{27}$，項数$m$）＋（初項0，公比$\frac{1}{27}$，項数$m$）＋（初項$(\frac{7}{3})^3$，公比$\frac{1}{27}$，項数$m$）
$=\frac{\frac{2}{3}(1-(\frac{1}{27})^m)}{1-\frac{1}{27}}+0+\frac{(\frac{7}{3})^3(1-(\frac{1}{27})^m)}{1-\frac{1}{27}}$
$=\frac{\frac{2}{3}(1-\frac{1}{3^{3m}})}{\frac{26}{27}}+\frac{\frac{7}{3}(1-\frac{1}{3^{3m}})}{\frac{26}{27}}$
$=\frac{25}{26}(1-\frac{1}{3^n})$

nが3の倍数から一つ少ない $n=3m-1$ のとき，$3m=n+1$で，
$S=\left((\frac{2}{3})^1+(\frac{0}{3})^2+(\frac{7}{3})^3\right)+\left((\frac{2}{3})^4+(\frac{0}{3})^5+(\frac{7}{3})^6\right)+\\ \cdots+\left((\frac{2}{3})^{n-5}+(\frac{0}{3})^{n-4}+(\frac{7}{3})^{n-3}\right)+\left((\frac{2}{3})^{n-2}+(\frac{0}{3})^{n-1}\right)$
$=\left((\frac{2}{3})^1+(\frac{2}{3})^4+(\frac{2}{3})^7+\cdots+(\frac{2}{3})^{3m-2}\right)\\ +\left((\frac{0}{3})^2+(\frac{0}{3})^5+(\frac{0}{3})^8+\cdots+(\frac{0}{3})^{3m-1}\right)\\ +\left((\frac{7}{3})^3+(\frac{7}{3})^6+(\frac{7}{3})^9+\cdots+(\frac{7}{3})^{3m-3}\right)$
＝（初項$\frac{2}{3}$，公比$\frac{1}{27}$，項数$m$）＋（初項0，公比$\frac{1}{27}$，項数$m$）＋（初項$(\frac{7}{3})^3$，公比$\frac{1}{27}$，項数$m-1$）
$=\frac{\frac{2}{3}(1-(\frac{1}{27})^m)}{1-\frac{1}{27}}+0+\frac{(\frac{7}{3})^3(1-(\frac{1}{27})^{m-1})}{1-\frac{1}{27}}$
$=\frac{1}{26}(25-\frac{23}{3^{n-1}})$


nが3の倍数から2つ少ない $n=n=3m-2$ のとき，$3m=n+2$より，同様にして，
$S=\frac{1}{26}(25-\frac{23}{3^{n}})$