수학적 귀납법 빈칸

[출제의도] 수학적 귀납법을 이용하여 명제를 증명한다. (ⅰ) $n = 1$일 때, $( \text{좌변} ) = 1$, $( \text{우변} ) = 1$이므로 (*)이 성립한다. (ⅱ) $n = m$일 때, (*)이 성립한다고 가정하면 $\displaystyle \sum\limits _{k = 1} ^{m} k \left\{ k + ( k + 1 ) + \cdots + m \right\} = \frac{m ( m + 1 ) ( m + 2 ) ( 3 m + 1 )}{24}$ 이므로 $\displaystyle \sum\limits _{k = 1} ^{m + 1} k \left\{ k + ( k + 1 ) + \cdots + ( m + 1 ) \right\}$ $\displaystyle = \sum\limits _{k=1} ^{m} k \left\{ k + ( k + 1 ) + \cdots + ( m + 1 ) \right\} + \square {( m + 1 ) ^{2}}$ $\displaystyle = \sum\limits _{k=1} ^{m} k \left\{ k + ( k + 1 ) + \cdots + m \right\} + \left( m + 1 \right) \sum\limits _{k=1} ^{m} k + \square {( m + 1 ) ^{2}}$ $\displaystyle = \sum\limits _{k = 1} ^{m} k \left\{ k + ( k + 1 ) + \cdots + m \right\} + {\square {\frac{m ( m + 1 ) ^{2}}{2}}} + {\square {( m + 1 ) ^{2}}}$ $\displaystyle = \frac{m ( m + 1 ) ( m + 2 ) ( 3 m + 1 )}{24} + {\square {\frac{m ( m + 1 ) ^{2}}{2}}} + {\square {( m + 1 ) ^{2}}}$ $\displaystyle = \frac{m ( m + 1 ) ( m + 2 ) ( 3 m + 1 )}{24} + \frac{\left( m + 1 \right) ^{2} \left( m + 2 \right)}{2}$ $\displaystyle = \frac{\left( m + 1 \right) \left( m + 2 \right) \left\{ m ( 3 m + 1 ) + 12 ( m + 1 ) \right\}}{24}$ $\displaystyle = \frac{\left( m + 1 \right) \left( m + 2 \right) \left( 3 m ^{2} + 13 m + 12 \right)}{24}$ $\displaystyle = \frac{( m + 1 ) ( m + 2 ) ( m + 3 ) ( 3 m + 4 )}{24}$ $\displaystyle = \frac{{( m + 1 ) \left\{ ( m + 1 ) + 1 \right\} \left\{ ( m + 1 ) + 2 \right\} \left\{ 3 ( m + 1 ) + 1 \right\}}}{24}$ 따라서 $n = m + 1$일 때도 (*)이 성립한다. (ⅰ), (ⅱ)에 의하여 모든 자연수 $n$에 대하여 (*)이 성립한다. $f ( m ) = ( m + 1 ) ^{2}$, $\displaystyle g ( m ) = \frac{m ( m + 1 ) ^{2}}{2}$이므로 $f ( 4 ) = ( 4 + 1 ) ^{2}$$= 25$, $\displaystyle g ( 2 ) = \frac{2 ( 2 + 1 ) ^{2}}{2} = 9$ 따라서 $f ( 4 ) + g ( 2 )$$= 25 + 9$$= 34$