数列 \(\{ a _ k \}\) を \(a _ k = k +\cos \left( \dfrac{k \pi}{6} \right)\) で定める. \(n\) を正の整数とする.
(1) \(\textstyle\sum\limits _ {k=1}^{12n} a _ k\) を求めよ.
(2) \(\textstyle\sum\limits _ {k=1}^{12n} {a _ k}^2\) を求めよ.
【 解 答 】
(1)
\(\cos ( x +\pi ) = \cos x\) , \(\cos \left( x -\dfrac{\pi}{2} \right) = -\cos x\) を用いれば \[ \textstyle\sum\limits _ {k=1}^{12n} \cos \left( \dfrac{k \pi}{6} \right) = n ( -1 +1 ) = 0 \] よって \[\begin{align} \textstyle\sum\limits _ {k=1}^{12n} a _ k & = \textstyle\sum\limits _ {k=1}^{12n} k \\ & = \underline{6n ( 12n+1 )} \end{align}\]
(2)
\[\begin{align} {a _ k}^2 & = k^2 +2 k \cos \dfrac{k \pi}{6} +\cos^2 \dfrac{k \pi}{6} \\ & = k^2 +2 \underline{k \cos \dfrac{k \pi}{6}} _ {[1]} +\dfrac{1}{2} \underline{\cos \dfrac{k \pi}{3}} _ {[2]} +\dfrac{1}{2} \end{align}\] ここで, 下線部 [1] について, 整数 \(m\) に対して \[\begin{align} \textstyle\sum\limits _ {k = 12m+1}^{12(m+1)} [1] & = \dfrac{\sqrt{3}}{2} \left\{ (12m+1) +(12m+11) -(12m+5) -(12m+7) \right\} \\ & \quad +\dfrac{1}{2} \left\{ (12m+2) +(12m+10) -(12m+4) -(12m+8) \right\} \\ & \qquad -1 \cdot (12m+6) +1 \cdot (12m+12) \\ & = 6 \end{align}\] なので \[ \textstyle\sum\limits _ {k=1}^{12n} [1] = 6n \] また, 下線部 [2] について, (1) と同様に考えれば \[ \textstyle\sum\limits _ {k=1}^{12n} [2] = n (-1+1) = 0 \] よって, 求める値は \[\begin{align} \textstyle\sum\limits _ {k=1}^{12n} {a _ k}^2 & = \textstyle\sum\limits _ {1}^{12n} k^2 +2 \cdot 6n +\dfrac{1}{2} \cdot 0 +\dfrac{1}{2} \cdot 12n \\ & = \dfrac{1}{6} \cdot 12n (12n+1) (24n+1) +12n +6n \\ & = 2n ( 288 n^2 +36n +1 +9 ) \\ & = \underline{4n ( 144n^2 +18n +5 )} \end{align}\]