上級者
数学B：数列
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ヒント：フィボナッチ数列ですが，あまり関係ありません．
問題
\(a_{n+2}=a_{n+1}+a_{n}\)，\(a_{1}=1\)，\(a_{2}=1\)のとき次のことを示せ．（オリジナル）
(1) \(n\)は4の倍数 \(\Leftrightarrow\) \(a_{n}\)は3の倍数
(2) \(a_{n}=n^{2}\)が成り立つとき，\(n\)を求めよ．
(3) \(\left[\displaystyle \frac{a_{n-1}}{a_{n}}\right]=0\)が\(n\geqq 3\)で成り立つことを示せ．
(4) \(a_{2n}=a_{n}(a_{n+1}+a_{n-1})\)が成り立つことを示せ．
(5) \(\displaystyle \sum_{k=1}^{n}a_{2k-1}=a_{2n}\)が成り立つことを示せ．
解答
(1)まず，数列のあまりに関してですが，これは周期的になります．
\(a_{1}=1,a_{2}=1,a_{3}=2,a_{4}=3,a_{5}=5,a_{6}=8,a_{7}=13,a_{8}=21,a_{9}=34,a_{10}=55,a_{11}=89\cdots\cdots\)
ここで，\(a_{n}\)を3で割った余りを\(b_{n}\)とすると，
\(b_{1}=1,b_{2}=1,b_{3}=2,b_{4}=0,b_{5}=2,b_{6}=2,b_{7}=1,b_{8}=0,b_{9}=1,b_{10}=1,b_{11}=2\cdots\cdots\)
よって周期8なので，見たように\(n\)が4の倍数 \(\Leftrightarrow\) \(a_{n}\)は3の倍数が成り立つ．
これは一応証明しておいたほうがいいかもしれない．
(2)具体的に調べてみよう．
\(a_{1}=1,a_{2}=1,a_{3}=2,a_{4}=3,a_{5}=5,a_{6}=8,a_{7}=13,a_{8}=21,a_{9}=34,a_{10}=55,a_{11}=89,a_{12}=144,a_{13}=233\cdots\)
なんとなく\(n=1,12\)しか成り立たないような．
\(n=1\)と\(n=12\)のときは上から明らかに成り立つ．
それ以外について，\(1<n<11\)については明らかに成り立たない．
\(n>12\)のとき\(a_{n}>n^{2}\)となりそう．これを数学的帰納法で証明する．
\(n=13,n=14\)のときにそれぞれ，\(a_{13}=233>13^{2},a_{14}=377>14^{2}\)が成り立つ．
ここで，\(n=k,n=k+1\)のとき\((k>13)\)，\(a_{k}>k^{2},~a_{k+1}>(k+1)^{2}\)が成り立つと仮定する．
すると\(a_{k+2}=a_{k+1}+a_{k}>(k+1)^{2}+k^{2}\)
ここで\((k+1)^{2}+k^{2}>(k+2)^{2}\)\(\Leftrightarrow\)\((k-3)(k+1)>0\)\(\Leftrightarrow\)\(k>3,k<-1\)
よって，\(a_{k+2}>(k+2)^{2}\)が成り立つ．
以上数学的帰納法より，\(a_{n}>n^{2}\)となる．
よって，\(n=1,12\)のみ成り立つ．
(3)これは単純である．\(\left[\displaystyle \frac{a_{n-1}}{a_{n}}\right]=\left[\frac{a_{n-1}}{a_{n-1}+a_{n-2}}\right]\)\(=\displaystyle \left[1-\frac{a_{n-2}}{a_{n-1}+a_{n-2}}\right]\)\((n\geqq 3)\)
ここで，\(0<\displaystyle \frac{a_{n-2}}{a_{n-1}+a_{n-2}}<1\)より，\(0<1-\displaystyle \frac{a_{n-2}}{a_{n-1}+a_{n-2}}<1\)
よって\(\left[1-\displaystyle \frac{a_{n-2}}{a_{n-1}+a_{n-2}}\right]=0\)で\(\left[\displaystyle \frac{a_{n-1}}{a_{n}}\right]=0\)
(4)ここでも数学的帰納法を使うとよい．
\(n=2,3\)のとき，\(a_{4}=a_{2}(a_{3}+a_{1})=1\times(2+1)=3\)，\(a_{6}=a_{3}(a_{4}+a_{2})=2\times(3+1)=8\)となり成り立つ．
ここで，\(n=k,k+1\)のとき，\(a_{2k}=a_{k}(a_{k+1}+a_{k-1})\)，\(a_{2(k+1)}=a_{k+1}(a_{k+2}+a_{k})\)が成り立つとすると，元の数列から，
\(a_{2(k+2)}=a_{2k+3}+a_{2k+2}=2a_{2k+2}+a_{2k+1}\)
ここで仮定より，\(a_{2(k+1)}=a_{k+1}(a_{k+2}+a_{k})\)\(\Leftrightarrow\)\(a_{2k+1}+a_{2k}=a_{k+1}(a_{k+2}+a_{k})\)\(\Leftrightarrow\)\(a_{2k+1}=a_{k+1}(a_{k+2}+a_{k})-a_{k}(a_{k+1}+a_{k-1})\)\(\Leftrightarrow\)\(a_{2k+1}=a_{k+2}a_{k+1}-a_{k}a_{k-1}\)
よって，\(a_{2(k+2)}=2a_{k+1}(a_{k+2}+a_{k})+a_{k+2}a_{k+1}-a_{k}a_{k-1}\)\(=2a_{k+1}(a_{k+2}+a_{k+2}-a_{k+1})+a_{k+2}a_{k+1}-(a_{k+2}-a_{k+1})(a_{k+1}-a_{k+2}+a_{k+1})\)\(=a_{k+2}(a_{k+2}+2a_{k+1})=a_{k+2}(a_{k+3}+a_{k+1})\)
よって\(n=k+2\)のときも成り立つ．
以上数学的帰納法より，\(a_{2n}=a_{n}(a_{n+1}+a_{n-1})\)が成り立つ．（これより\(a_{2n+1}=a_{n+2}a_{n+1}-a_{n}a_{n-1}\)も成り立つことがわかる．）
(5) (4)の過程からわかる．
\(\displaystyle \sum_{k=1}^{n}a_{2k-1}=a_{1}+a_{3}+\sum_{k=3}^{n}a_{k+1}a_{k}-a_{k-1}a_{k-2}\)\(=a_{1}+a_{2}+a_{4}a_{3}-a_{2}a_{1}+a_{5}a_{4}-a_{3}a_{2}+\cdots+a_{n+1}a_{n}-a_{n-1}a_{n-2}\)\(=a_{1}+a_{3}-a_{2}a_{1}-a_{3}a_{2}+a_{n}a_{n-1}+a_{n+1}a_{n}=a_{2n}\)
\(\star\)ほかにもいろいろな性質がある．
1,\(\displaystyle a_{n}=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right\}\)
2,\(\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n}}{a_{n-1}}=\frac{1+\sqrt{5}}{2}\)
3,\(\displaystyle \sum_{k=1}^{n}a_{k}=a_{n+2}-1\)
4,\(\displaystyle \sum_{k=1}^{n}a_{2k}=a_{2n+1}-1\)
5,\(\displaystyle \sum_{k=1}^{n}a_{k}^{2}=a_{n}a_{n+1}\)
6,\(a_{n-1}a_{n+1}-a_{n}^{2}=(-1)^{n}\)
1,の証明
三項間漸化式
\(a_{n+2}=a_{n+1}+a_{n}\)で，特性方程式 \(x^{2}=x+1\)の解を\(s,t\)とする．\((s<t)\)
すると，\(a_{n+2}-sa_{n+1}=t(a_{n+1}-sa_{n})\)
\(a_{n+2}-ta_{n+1}=s(a_{n+1}-ta_{n})\)となり，それぞれ解いて，
\(a_{n+1}-sa_{n}=t^{n-1}(a_{2}-sa_{1})\)
\(a_{n+1}-ta_{n}=s^{n-1}(a_{2}-ta_{1})\)
これから\(a_{n+1}\)を消すと，\((t-s)a_{n}=t^{n}-s^{n}\)\(\Leftrightarrow\)\(\displaystyle a_{n}=\frac{1}{\sqrt{5}}\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right\}\) (\(a_{2}-sa_{1}=t\)，\(a_{2}-ta_{1}=s\)，\(t-s=\sqrt{5})\))
2,の証明
特殊な二項間漸化式
\(\displaystyle \frac{a_{n}}{a_{n-1}}=A_{n}\)とすると，\(\displaystyle A_{n+1}=\frac{a_{n}+a_{n-1}}{a_{n}}\)\(=\displaystyle 1+\frac{1}{A_{n}}\)\(=\displaystyle \frac{A_{n}+1}{A_{n}}\)
ここで，一般的に，\(a_{n}=\displaystyle \frac{sa_{n-1}+t}{ua_{n-1}+v}\)のとき，特性方程式\(x=\displaystyle \frac{sx+t}{ux+v}\)の解を\(p,q\)とすると，\(b_{n}=\displaystyle \frac{a_{n}-p}{a_{n}-q}\)（等比数列）もしくは \(b_{n}=\displaystyle \frac{1}{a_{n}-p}\)（\(a_{n+1}=xa_{n}+y\)の形）となる．
よって，この解を\(s,t\)（さっきと同じ）とおくと，\(b_{n+1}=\displaystyle \frac{A_{n+1}-t}{A_{n+1}-s}\)\(=\displaystyle \frac{\frac{A_{n}+1}{A_{n}}-t}{\frac{A_{n}+1}{A_{n}}-s}\)\(=\displaystyle \frac{(1-t)A_{n}+1}{(1-s)A_{n}+1}\)\(=\displaystyle \frac{s}{t}\left(\frac{A_{n}+\frac{1}{s}}{A_{n}+\frac{1}{t}}\right)\)\(=\displaystyle \frac{s}{t}\left(\frac{A_{n}-t}{A_{n}-s}\right)\)\(=\displaystyle \frac{s}{t}b_{n} (1-s=t,~st=-1)\)\((n\geqq 2)\)
よって，\(\displaystyle b_{n}=\left(\frac{s}{t}\right)^{n-2}\left(\frac{A_{2}-t}{A_{2}-s}\right)\)\(=\displaystyle \left(\frac{s}{t}\right)^{n-1}\)\(\Leftrightarrow\)\(\displaystyle \frac{A_{n}-t}{A_{n}-s}\)\(=\displaystyle \left(\frac{s}{t}\right)^{n-1}\)\(\Leftrightarrow\)\(\displaystyle A_{n}=\frac{t-su^{n-1}}{1-u^{n-1}}\) (\(\displaystyle u=\frac{s}{t}\), \(0<u<1\))
ゆえに\(\displaystyle \lim_{n\rightarrow\infty}A_{n}\)\(=\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n}}{a_{n-1}}\)\(=\displaystyle t=\frac{1+\sqrt{5}}{2}\)
3,の証明 1,を使おう．
\(\displaystyle \sum_{k=1}^{n}a_{k}=\frac{1}{\sqrt{5}}\sum_{k=1}^{n}(t^{k}-s^{k})\)\(=\displaystyle \frac{1}{\sqrt{5}}\left(t\frac{1-t^{n}}{1-t}-s\frac{1-s^{n}}{1-s}\right)\)\(=\displaystyle -\frac{1}{\sqrt{5}}(t^{2}-t^{n+2}-s^{2}+s^{n+2})\)\(=\displaystyle \frac{1}{\sqrt{5}}(t^{n+2}-s^{n+2})-1\)\(=a_{n+2}-1\) \(\displaystyle (t^{2}-s^{2}=\sqrt{5})\)
4,の証明(4)，(5)と3,を使おう．
\(\displaystyle \sum_{k=1}^{2n}a_{k}=\sum_{k=1}^{n}a_{2k}+\sum_{k=1}^{n}a_{2k-1}\)\(\Leftrightarrow\)\(\displaystyle \sum_{k=1}^{n}a_{2k}=\sum_{k=1}^{2n}a_{k}-\sum_{k=1}^{n}a_{2k-1}\)\(=a_{2n+2}-1-a_{2n}\)\(=a_{n+1}(a_{n+2}+a_{n})-a_{n}(a_{n+1}+a_{n-1})-1\)\(=a_{n+2}a_{n+1}-a_{n}a_{n-1}-1=a_{2n+1}-1\)
5,の証明
\(\displaystyle \sum_{k=1}^{n}a_{k}^{2}=\frac{1}{5}\sum_{k=1}^{n}(t^{k}-s^{k})^{2}\)\(=\displaystyle \frac{1}{5}\sum_{k=1}^{n}\left(t^{2k}-2(ts)^{k}+s^{2k}\right)\)\(=\displaystyle \frac{1}{5}(t^{2}\frac{1-t^{2n}}{1-t^{2}}+s^{2}\frac{1-s^{2n}}{1-s^{2}}+1-(-1)^{n})\) \((1-t^{2}=-t,~1-s^{2}=-s,~st=-1)\)\(=\displaystyle \frac{1}{5}(t^{2n+1}+s^{2n+1}-t-s+1-(-1)^{n})\)\(=\displaystyle \frac{1}{5}(t^{2n+1}+s^{2n+1}-(-1)^{n})\) \((t+s=1)\)
\(a_{n}a_{n+1}=\displaystyle \frac{1}{5}(t^{n}-s^{n})(t^{n+1}-s^{n+1})\)\(=\displaystyle \frac{1}{5}(t^{2n+1}+s^{2n+1}-(st)^{n}(s+t))=\frac{1}{5}(t^{2n+1}+s^{2n+1}-(-1)^{n})\)
よって\(\displaystyle \sum_{k=1}^{n}a_{k}^{2}=a_{n}a_{n+1}\)
6,の証明
\(a_{n-1}a_{n+1}-a_{n}^{2}=a_{n-1}(a_{n}+a_{n-1})-(a_{n-1}+a_{n-2})^{2}\)\(=a_{n}a_{n-1}+a_{n-1}^{2}-a_{n-1}^{2}-2a_{n-1}a_{n-2}-a_{n-2}^{2}\)\(=a_{n-1}(a_{n}-2a_{n-2})=a_{n-3}a_{n-1}-a_{n-2}^{2}\) \((a_{n}-2a_{n-2}=a_{n-1}+a_{n-2}-2a_{n-2}=a_{n-2}+a_{n-3}+a_{n-2}-2a_{n-2}=a_{n-3})\)
\(b_{n}=a_{n-1}a_{n+1}-a_{n}^{2}\)とおけば，\(b_{n+2}=b_{n}\)\((n\geqq 2)\)
ここで，\(n\)が奇数のとき，\(b_{n}=\cdots=b_{3}=a_{2}a_{4}-a_{3}^{2}=-1\)
\(n\)が偶数のとき，\(b_{n}=\cdots=b_{2}=a_{1}a_{3}-a_{2}^{2}=1\)
以上より，\(b_{n}=a_{n-1}a_{n+1}-a_{n}^{2}\)\(=(-1)^{n}\)