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

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

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

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

$$\begin{equation} \left\{ \begin{array}{l} a - b = 0 \\ 3a - 2b = 18 \end{array} \right. \end{equation}$$

$$\begin{equation} a = 18,~b = 18 \end{equation}$$

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

$$\begin{align} E[Y(Y + 1)] &= \sum_{k=1}^{\infty}k(k+1)\,P(Y = k) \\ &= \lim_{n \to \infty}\sum_{k=1}^{n}k(k+1)\,\frac{18}{k(k+1)(k+2)(k+3)} \\ &= \lim_{n \to \infty}\sum_{k=1}^{n}\frac{18}{(k+2)(k+3)} \\ &= \lim_{n \to \infty}\sum_{k=1}^{n}\left\{\frac{18}{k+2} - \frac{18}{k+3}\right\} \\ &= \lim_{n \to \infty}\left\{ \left(\frac{18}{3} - \frac{18}{4}\right) + \left(\frac{18}{4} - \frac{18}{5}\right) + \left(\frac{18}{5} - \frac{18}{6}\right) \right. \\ &~~~~ + \cdots + \left.\left(\frac{18}{n} - \frac{18}{n+1}\right) \right\} \\ &= \lim_{n \to \infty}\left\{\frac{18}{3} - \frac{18}{n+1}\right\} \\ &= \frac{18}{3} \\ &= 6 \end{align}$$

$$\begin{align} V[Y] &= E[Y(Y + 1)] - E[Y] - \left(E[Y]\right)^2 \\ &= 6 - \frac{3}{2} - \left(\frac{3}{2}\right)^2 \\ &= \frac{9}{4} \end{align}$$