거듭제곱 합 등식 귀납법

[출제의도] 수학적 귀납법 추론하기 (1) $n = 1$일 때, $( \text{좌변} ) = 6 \times 1 \times 1 ^{2} = 6$이고, $( \text{우변} ) = 5 \times 1 ^{4} + 1 ^{2} = 6$이므로 (*)이 성립한다. (2) $n = m$일 때, (*)이 성립한다고 가정하면 $\displaystyle 6 \left( \sum\limits _{k = 1} ^{m} k \right) \left( \sum\limits _{k = 1} ^{m} k ^{2} \right) = 5 \sum\limits _{k = 1} ^{m} k ^{4} + \sum\limits _{k = 1} ^{m} k ^{2}$ 이다. $n = m + 1$일 때, (*)이 성립함을 보이자. $\displaystyle 6 \left( \sum\limits _{k = 1} ^{m + 1} k \right) \left( \sum\limits _{k = 1} ^{m + 1} k ^{2} \right)$ $\displaystyle = 6 \left\{ \sum\limits _{k = 1} ^{m} k + \left( \square {( m + 1 )} \right) \right\} \left\{ \sum\limits _{k = 1} ^{m} k ^{2} + \left( m + 1 \right) ^{2} \right\}$ $\displaystyle = 6 \left\{ \left( \sum\limits _{k = 1} ^{m} k \right) \left( \sum\limits _{k = 1} ^{m} k ^{2} \right) + \left( m + 1 \right) \sum\limits _{k = 1} ^{m} k ^{2} + \left( m + 1 \right) ^{2} \sum\limits _{k = 1} ^{m} k + \left( m + 1 \right) ^{3} \right\}$ $\displaystyle = 6 \left( \sum\limits _{k = 1} ^{m} k \right) \left( \sum\limits _{k = 1} ^{m} k ^{2} \right)$ $\displaystyle + \left( {\square {( m + 1 )}} \right) \times \left\{ 6 \sum\limits _{k = 1} ^{m} k ^{2} + 6 \left( m + 1 \right) \sum\limits _{k = 1} ^{m} k + 6 \left( m + 1 \right) ^{2} \right\}$ $\displaystyle = 6 \left( \sum\limits _{k = 1} ^{m} k \right) \left( \sum\limits _{k = 1} ^{m} k ^{2} \right)$ $+ \left( m + 1 \right) \times \left\{ m \left( m + 1 \right) \left( 2 m + 1 \right) + 3 m \left( m + 1 \right) ^{2} + 6 \left( m + 1 \right) ^{2} \right\}$ $\displaystyle = 6 \left( \sum\limits _{k = 1} ^{m} k \right) \left( \sum\limits _{k = 1} ^{m} k ^{2} \right)$ $+ \left( m + 1 \right) ^{2} \times \left\{ m \left( 2 m + 1 \right) + 3 m \left( m + 1 \right) + 6 \left( m + 1 \right) \right\}$ $\displaystyle = 6 \left( \sum\limits _{k = 1} ^{m} k \right) \left( \sum\limits _{k = 1} ^{m} k ^{2} \right)$$+ \left( m + 1 \right) ^{2} \times ( {\square {5 m ^{2} + 10 m + 6}} )$ $\displaystyle = 6 \left( \sum\limits _{k = 1} ^{m} k \right) \left( \sum\limits _{k = 1} ^{m} k ^{2} \right)$$+ \left( m + 1 \right) ^{2} \times \left\{ 5 \left( m + 1 \right) ^{2} + 1 \right\}$ $\displaystyle = 6 \left( \sum\limits _{k = 1} ^{m} k \right) \left( \sum\limits _{k = 1} ^{m} k ^{2} \right)$$+ 5 \left( m + 1 \right) ^{4} + \left( m + 1 \right) ^{2}$ $\displaystyle = 5 \sum\limits _{k = 1} ^{m} k ^{4} + \sum\limits _{k = 1} ^{m} k ^{2} + 5 \left( m + 1 \right) ^{4} + \left( m + 1 \right) ^{2}$ $\displaystyle = 5 \sum\limits _{k = 1} ^{m} k ^{4} + 5 \left( m + 1 \right) ^{4} + \sum\limits _{k = 1} ^{m} k ^{2} + \left( m + 1 \right) ^{2}$ $\displaystyle = 5 \sum\limits _{k = 1} ^{m + 1} k ^{4} + \sum\limits _{k = 1} ^{m + 1} k ^{2}$ 그러므로 $n = m + 1$일 때도 (*)이 성립한다. 따라서 모든 자연수 $n$에 대하여 (*)이 성립한다. $f ( m ) = m + 1$, $g ( m ) = 5 m ^{2} + 10 m + 6$이므로 $\displaystyle \frac{g ( 10 )}{f ( 5 )} = \frac{606}{6} = 101$이다.