数列 \(\{ a _ n \}\) は \[ a _ 1 = 5 , \quad {a _ 1}^2 +{a _ 2}^2 + \cdots +{a _ n}^2 = \dfrac{2}{3} a _ n a _ {n+1} \quad ( n = 1, 2, 3, \cdots ) \] をみたすとする. 次の問いに答えよ.
(1) \(a _ 2 , a _ 3\) を求めよ.
(2) \(a _ {n+2}\) を \(a _ n , a _ {n+1}\) を用いて表せ.
(3) 一般項 \(a _ n\) を求めよ.
【 解 答 】
(1)
与えられた漸化式について \(n = 1\) とすれば \[\begin{align} 5^2 & = \dfrac{2}{3} \cdot 5 a_2 \\ \text{∴} \quad a_2 & = \underline{\dfrac{15}{2}} \end{align}\] また, \(n = 2\) とすれば \[\begin{align} 5^2 +\left( \dfrac{15}{2} \right)^2 & = \dfrac{2}{3} \cdot \dfrac{15}{2} a_3 \\ \text{∴} \quad a_3 & = \underline{\dfrac{65}{4}} \end{align}\]
(2)
\[\begin{align} {a_1}^2 +{a_2}^2 +\cdots +{a_n}^2 & = \dfrac{2}{3} a_n a _ {n+1} \quad ... [1] \\ {a_1}^2 +{a_2}^2 +\cdots +{a_n}^2 +{a _ {n+1}}^2 & = \dfrac{2}{3} a _ {n+1} a _ {n+2} \quad ... [2] \end{align}\] \([2] -[1]\) より \[ {a _ {n+1}}^2 = \dfrac{2}{3} a _ {n+1} \left( a _ {n+2} -a_n \right) \] \(( [1] \text{の左辺} ) \neq 0\) なので \(a _ {n+1} \neq 0\) だから \[\begin{align} a _ {n+1} & = \dfrac{2}{3} \left( a _ {n+2} -a_n \right) \\ \text{∴} \quad a _ {n+2} & = \underline{\dfrac{3}{2} a _ {n+1} +a_n} \quad ... [3] \end{align}\]
(3)
[3] を変形すると \[ a _ {n+2} +\dfrac{1}{2} a _ {n+1} = 2 \left( a _ {n+1} +\dfrac{1}{2} a_n \right) \] 数列 \(\left\{ a _ {n+1} +\dfrac{1}{2} a_n \right\}\) は, 初項 \(a_2 +\dfrac{1}{2} a_1 = \dfrac{15}{2} +\dfrac{5}{2} = 10\) , 公比 \(2\) の等比数列なので \[\begin{align} a _ {n+1} +\dfrac{1}{2} a_n & = 10 \cdot 2^{n-1} = 5 \cdot 2^n \\ \text{∴} \quad a _ {n+1} -2^{n+2} & = -\dfrac{1}{2} \left( a_n -2^{n+1} \right) \end{align}\] 数列 \(\left\{ a_n -2^{n+1} \right\}\) は, 初項 \(a_1 -2^2 = 1\) , 公比 \(-\dfrac{1}{2}\) の等比数列なので \[\begin{align} a_n -2^{n+1} & = \left( -\dfrac{1}{2} \right)^{n-1} \\ \text{∴} \quad a_n & = \underline{2^{n+1} +\left( -\dfrac{1}{2} \right)^{n-1}} \end{align}\]