2011年7月26日火曜日

数列

\frac{23}{111}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)+\\ \hspace{20}\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}{(3^3)^m})}{\frac{26}{27}}+\frac{\frac{7}{3}(1-\frac{1}{(3^3)^m})}{\frac{26}{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)+\\ \hspace{20}\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}})

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