統計検定 1級 過去問 解答/解答例と解説
2015年11月29日 (日) 試験

$$\begin{equation} \frac{18}{(k+1)(k+2)(k+3)} = \frac{a}{(k+1)(k+2)} - \frac{b}{(k+2)(k+3)} \end{equation}$$

$$\begin{align} \frac{18}{(k+1)(k+2)(k+3)} &= \frac{a(k+3) - b(k+1)}{(k+1)(k+2)(k+3)} \\ &= \frac{ak+3a - bk - b}{(k+1)(k+2)(k+3)} \\ &= \frac{(a - b)k + (3a - b)}{(k+1)(k+2)(k+3)} \\ \end{align}$$

$$\begin{equation} (a - b)k + (3a - b) = 18 \end{equation}$$

$$\begin{equation} \left\{ \begin{array}{l} a - b = 0 \\ 3a - b = 18 \end{array} \right. \end{equation}$$

$$\begin{equation} a = 9,~b = 9 \end{equation}$$

$$\begin{equation} \frac{18}{(k+1)(k+2)(k+3)} = \frac{9}{(k+1)(k+2)} - \frac{9}{(k+2)(k+3)} \end{equation}$$

$$\begin{align} E[Y] &= \sum_{k=1}^{\infty}k\,P(Y = k) \\ &= \lim_{n \to \infty}\sum_{k=1}^{n}k\,\frac{18}{k(k+1)(k+2)(k+3)} \\ &= \lim_{n \to \infty}\sum_{k=1}^{n}\frac{18}{(k+1)(k+2)(k+3)} \\ &= \lim_{n \to \infty}\sum_{k=1}^{n}\left\{\frac{9}{(k+1)(k+2)} - \frac{9}{(k+2)(k+3)}\right\} \\ &= \lim_{n \to \infty}\left\{\left(\frac{9}{6} - \frac{9}{12}\right) + \left(\frac{9}{12} - \frac{9}{20}\right) + \left(\frac{9}{20} - \frac{9}{30}\right)\right. \\ &~~~~ + \cdots + \left. \left(\frac{9}{(n+1)(n+2)} - \frac{9}{(n+2)(n+3)}\right)\right\} \\ &= \lim_{n \to \infty}\left\{\frac{9}{6} - \frac{9}{(n+2)(n+3)}\right\} \\ &= \frac{9}{6} \\ &= \frac{3}{2} \\ \end{align}$$