귀납법 시그마 증명

[출제의도] 수학적 귀납법 추론하기 (ⅰ) $n = 1$일 때, (좌변)$\displaystyle = \sum\limits _{k = 1} ^{1} \left( a _{k} a _{k + 1} \right) ^{2} = {\square {\frac{1}{4}}}$, (우변)$\displaystyle = \sum\limits _{k = 1} ^{1} \left( a _{k} \right) ^{2} + \sum\limits _{k = 1} ^{1} \left( a _{k + 1} \right) ^{2} + 2 \left( a _{2} - 1 \right) = {\square {\frac{1}{4}}}$이므로 (*)이 성립한다. (ⅱ) $n = m \left( m \geq 1 \right)$일 때, (*)이 성립한다고 가정하면 $\displaystyle \sum\limits _{k = 1} ^{m} \left( a _{k} a _{k + 1} \right) ^{2} = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{2} + \sum\limits _{k = 1} ^{m} \left( a _{k + 1} \right) ^{2} + 2 \left( a _{m + 1} - 1 \right)$ 이므로 $\displaystyle \sum\limits _{k = 1} ^{m + 1} \left( a _{k} a _{k + 1} \right) ^{2}$$\displaystyle = \sum\limits _{k = 1} ^{m} \left( a _{k} a _{k + 1} \right) ^{2} + \left( a _{m + 1} a _{m + 2} \right) ^{2}$ $\displaystyle = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{2} + \sum\limits _{k = 1} ^{m} \left( a _{k + 1} \right) ^{2} + 2 \left( a _{m + 1} - 1 \right) + \left( a _{m + 1} a _{m + 2} \right) ^{2}$ $\displaystyle = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{2} + \sum\limits _{k = 1} ^{m} \left( a _{k + 1} \right) ^{2} + 2 \left( \frac{1}{m + 1} - 1 \right) + \left\{ \frac{1}{\left( m + 1 \right) \left( m + 2 \right)} \right\} ^{2}$ $\displaystyle = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{2} + \sum\limits _{k = 1} ^{m} \left( a _{k + 1} \right) ^{2} + 2 \left( \frac{1}{m + 1} - 1 \right) + \left( \frac{1}{m + 1} - \frac{1}{m + 2} \right) ^{2}$ $\displaystyle = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{2} + \sum\limits _{k = 1} ^{m} \left( a _{k + 1} \right) ^{2} + 2 \left\{ \frac{1}{m + 1} - 1 - \frac{1}{\left( m + 1 \right) \left( m + 2 \right)} \right\}$ $\displaystyle + \left( \frac{1}{m + 1} \right) ^{2} + \left( \frac{1}{m + 2} \right) ^{2}$ $\displaystyle = \sum\limits _{k = 1} ^{m + 1} \left( a _{k} \right) ^{2} + \sum\limits _{k = 1} ^{m + 1} \left( a _{k + 1} \right) ^{2} + 2 \left\{ \frac{1}{m + 1} - 1 - {\square {\frac{1}{\left( m + 1 \right) \left( m + 2 \right) ^{}}}} \right\}$ $\displaystyle = \sum\limits _{k = 1} ^{m + 1} \left( a _{k} \right) ^{2} + \sum\limits _{k = 1} ^{m + 1} \left( a _{k + 1} \right) ^{2} + 2 \left( \frac{1}{m + 2} - 1 \right)$ $\displaystyle = \sum\limits _{k = 1} ^{m + 1} \left( a _{k} \right) ^{2} + \sum\limits _{k = 1} ^{m + 1} \left( a _{k + 1} \right) ^{2} + 2 \left( a _{m + 2} - 1 \right)$ 이다. 따라서 $n = m + 1$일 때에도 (*)이 성립한다. (ⅰ), (ⅱ)에 의하여 모든 자연수 $n$에 대하여 (*)이 성립한다. 이 과정에서 $\displaystyle p = \frac{1}{4}$, $\displaystyle f ( m ) = \frac{1}{\left( m + 1 \right) \left( m + 2 \right)}$이므로 $\displaystyle \frac{p}{f ( 14 )} = 60$이다.