数列 \(\{ a _ n \}\) を, \[\begin{align} a _ 1 & = 1 , \\ (n+3) a _ {n+1} -n a _ n & = \dfrac{1}{n+1} -\dfrac{1}{n+2} \quad ( n = 1 , 2 , 3 , \cdots ) \end{align}\] によって定める.
(1) \(b _ n = n(n+1)(n+2) a _ n \ ( n = 1 , 2 , 3 , \cdots )\) によって定まる数列 \(\{ b _ n \}\) の一般項を求めよ.
(2) 等式 \[ p(n+1)(n+2) +qn(n+2) +rn(n+1) = b _ n \quad ( n = 1 , 2 , 3 , \cdots ) \] 成り立つように, 定数 \(p , q , r\) の値を定めよ.
(3) \(\textstyle\sum\limits _ {k=1}^n a _ k\) を \(n\) の式で表せ.
【 解 答 】
(1)
漸化式の両辺に \((n+1)(n+2)\) をかけると \[\begin{align} (n+1)(n+2)(n+3) a _ {n+1} -n(n+1)(n+2) a _ n & = (n+2) -(n+1) \\ \text{∴} \quad b _ {n+1} -b _ n & = 1 \end{align}\] したがって, \(\{ b _ n \}\) は, 初項 \(b _ 1 = 1 \cdot 2 \cdot 3 a _ 1 = 6\) , 公差 \(1\) の等差数列で \[ b _ n = 6 +1 \cdot (n-1) = \underline{n-5} \]
(2)
\[\begin{align} p (n+1) & (n+2) +qn(n+2) +rn(n+1) \\ & = p (n^2+3n+2) +q (n^2+2n) r (n^2+n) \\ & = (p+q+r) n^2 +(3p+2q+r)n +2p \end{align}\] これと (1) の結果を比較すれば, \[\begin{gather} \left\{ \begin{array}{l} p+q+r=0 \\ 3p+2q+r=1 \\ 2p=5 \end{array} \right. \\ \text{∴} \quad p =\underline{\dfrac{5}{2}} , \ q = \underline{-4} , \ r = \underline{\dfrac{3}{2}} \end{gather}\]
(3)
(2) の結果について, 両辺を \(n(n+1)(n+2)\) で割れば \[\begin{align} a _ n & = \dfrac{5}{2n} -\dfrac{4}{n+1} +\dfrac{3}{2(n+2)} \\ & = \dfrac{5}{2} \left( \dfrac{1}{n} -\dfrac{1}{n+1} \right) -\dfrac{3}{2} \left( \dfrac{1}{n+1} -\dfrac{1}{n+2} \right) \end{align}\] ゆえに \[\begin{align} \textstyle\sum\limits _ {k=1}^n a _ k & = \dfrac{5}{2} \left( 1 -\dfrac{1}{n+1} \right) -\dfrac{3}{2} \left( \dfrac{1}{2} -\dfrac{1}{n+2} \right) \\ & = \dfrac{7}{4} -\dfrac{5}{2(n+1)} +\dfrac{3}{2(n+2)} \\ & = \dfrac{7(n+1)(n+2) -10(n+2) +6(n+1)}{4(n+1)(n+2)} \\ & = \underline{\dfrac{n(7n+17)}{4(n+1)(n+2)}} \end{align}\]