複素数 \(\alpha = \dfrac{-1 +\sqrt{3} i}{2}\) に対して, \[ S _ n = \textstyle\sum\limits _ {k=1}^n {\alpha}^{k-1} , \ T _ n = \textstyle\sum\limits _ {k=1}^n k {\alpha}^{k-1} \quad ( n = 1, 2, \cdots ) \] とおく. ただし, \({\alpha}^0 = 1\) とする. 次の問に答えよ.
(1) \(S _ {3m} \ ( n = 1, 2, \cdots )\) を求めよ.
(2) \(T _ {3m} \ ( n = 1, 2, \cdots )\) を求めよ.
(3) \(T _ {2014}\) を求めよ.
【 解 答 】
\(\alpha\) は \(1\) の \(3\) 乗根なので \[ {\alpha}^3 = 1 \] また, これを変形して \[\begin{align} ( \alpha -1 ) ( {\alpha}^2 +\alpha +1 ) & = 0 \\ \text{∴} \quad {\alpha}^2 +\alpha +1 & = 0 \quad ( \ \text{∵} \ \alpha \neq 1 \ ) \end{align}\]
(1)
\[ S _ {3m} = m ( {\alpha}^2 +\alpha +1 ) = \underline{0} \]
(2)
\[\begin{align} T _ {3m} & = 1 ( 1 +4 + \cdots +3m-2 ) \\ & \quad \alpha ( 2 +5 + \cdots + 3m-1 ) \\ & \qquad {\alpha}^2 ( 3 +6 + \cdots +3m ) \\ & = \dfrac{m ( 3m-1 )}{2} +\dfrac{m ( 3m+1 ) \alpha}{2} +\dfrac{m ( 3m+3 ) {\alpha}^2}{2} \\ & = \dfrac{m}{2} \left\{ 3m-1 +( 3m+1 ) \dfrac{-1 +\sqrt{3} i}{2} +( 3m+3 ) \dfrac{-1 -\sqrt{3} i}{2} \right\} \\ & = \underline{-\dfrac{m \left( 3 +\sqrt{3} i \right)}{2}} \end{align}\]
(3)
\(2014 = 3 \cdot 671 +1\) なので \[\begin{align} T _ {2014} & = T _ {3 \cdot 671} +2014 {\alpha}^{2013} \\ & = -\dfrac{671 \left( 3 +\sqrt{3} i \right)}{2} +2014 \\ & = \underline{\dfrac{2015 -671 \sqrt{3} i}{2}} \end{align}\]