수학적 귀납법 빈칸

[출제의도] 수학적 귀납법을 이용하여 증명과정 추론하기 (i) $n = 1$일 때 (좌변)$\displaystyle = \left( \sum\limits _{k = 1} ^{1} a _{k} \right) ^{2} = \square {4}$, (우변)$\displaystyle = \sum\limits _{k = 1} ^{1} \left( a _{k} \right) ^{3} - 2 \sum\limits _{k = 1} ^{1} a _{k} = \square {4}$이므로 (*)이 성립한다. (ii) $n = m ( m \geq 1 )$일 때, (*)이 성립한다고 가정하면 $\displaystyle \left( \sum\limits _{k = 1} ^{m} a _{k} \right) ^{2} = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{3} - 2 \sum\limits _{k = 1} ^{m} a _{k}$이므로 $\displaystyle \left( \sum\limits _{k = 1} ^{m + 1} a _{k} \right) ^{2} = \left( \sum\limits _{k = 1} ^{m} a _{k} + a _{m + 1} \right) ^{2}$ $\displaystyle = \left( \sum\limits _{k = 1} ^{m} a _{k} \right) ^{2} + 2 \left( \sum\limits _{k = 1} ^{m} a _{k} \right) a _{m + 1} + \left( a _{m + 1} \right) ^{2}$ $\displaystyle = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{3} - 2 \sum\limits _{k = 1} ^{m} a _{k} + 2 \left( \sum\limits _{k = 1} ^{m} a _{k} \right) a _{m + 1} + \left( a _{m + 1} \right) ^{2}$ $\displaystyle = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{3} + \left( {\square {2 m + 2}} \right) \sum\limits _{k = 1} ^{m} a _{k} + \left( a _{m + 1} \right) ^{2}$ $\displaystyle = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{3} + {m ^{3} + 5 m ^{2} + 7 m + 4}$ $\displaystyle = \sum\limits _{k = 1} ^{m} \left( a _{k} \right) ^{3} + ( a _{m + 1} ) ^{3} - \left( {m ^{2} + 5 m + 4} \right. )$ $\displaystyle = \sum\limits _{k = 1} ^{m + 1} \left( a _{k} \right) ^{3} - 2 \sum\limits _{k = 1} ^{m + 1} a _{k}$ 이다. 따라서 $n = m + 1$일 때에도 (*)이 성립한다. (i), (ii)에 의하여 모든 자연수 $n$에 대하여 (*)이 성립한다. $p = 4 ,$ $f \left( m \right) = 2 m + 2$이므로 $f \left( 4 \right) = 10$