자연계열 2번

[2-1 풀이]

$\displaystyle f ( x ) = x^{n + 1} - x^{n + 2}$이므로

$\displaystyle f^{\prime} ( x ) = ( n + 1 ) x^{n} - ( n + 2 ) x^{n + 1} = x^{n} \left\{ n + 1 - ( n + 2 ) x \right\}$

점 $\displaystyle ( a , f ( a ) )$에 있어서의 접선의 방정식은 $\displaystyle y = f^{\prime} ( a ) ( x - a ) + f ( a )$

이것이 원점을 통과하므로

$\displaystyle 0 = - f^{\prime} ( a ) a + f ( a )$

$\displaystyle = - a^{n + 1} \left\{ n + 1 - \left( n + 2 \right) a \right\} + a^{n + 1} ( 1 - a )$

$\displaystyle = a^{n + 1} \left\{ \left( n + 1 \right) a - n \right\}$

$\displaystyle a > 0$이므로 $\displaystyle a = \frac{{n}}{n + 1}$

[2-2 풀이]

$\displaystyle 0 \leq x \leq \frac{n}{n + 1} < 1$에서 $\displaystyle f ( x ) = x^{n + 1} ( 1 - x ) > 0$

$\displaystyle A_{n} = \int_{0}^{\frac{{n}}{n + 1}} {f ( x ) d x} = \int_{0}^{\frac{{n}}{n + 1}} {( x^{n + 1} - x^{n + 2} )} d x$

$\displaystyle = \Biggl[ \frac{{x^{n + 2}}}{n + 2} - \frac{{x^{n + 3}}}{n + 3} \Biggr]_{0}^{\frac{{n}}{n + 1}} = \frac{{1}}{n + 2} \left( \frac{{n}}{n + 1} \right)^{n + 2} - \frac{{1}}{n + 3} \left( \frac{{n}}{n + 1} \right)^{n + 3}$

$\displaystyle = \left( \frac{{n}}{n + 1} \right)^{n + 2} \left( \frac{{1}}{n + 2} - \frac{{n}}{( n + 1 ) ( n + 3 )} \right)= \left( \frac{{n}}{n + 1} \right)^{n + 2} \frac{{2 n + 3}}{( n + 1 ) ( n + 2 ) ( n + 3 )}$

$\displaystyle ( n + 2 ) ( n + 3 ) A_{n} = \left( \frac{{n}}{n + 1} \right)^{n + 2} \frac{{2 n + 3}}{n + 1}= \left( \frac{{1}}{1 + \frac{{1}}{n}} \right)^{n + 2} \frac{{2 + \frac{{3}}{n}}}{1 + \frac{{1}}{n}}= \frac{{1}}{\left( 1 + \frac{{1}}{n} \right)^{n}} \bullet \frac{{2 + \frac{{3}}{n}}}{\left( 1 + \frac{{1}}{n} \right)^{3}}$

$\displaystyle \lim_{n \to \infty} {( n + 2 ) ( n + 3 ) A_{n}} = \lim_{n\to\infty} {\frac{{1}}{\left( 1 + \frac{{1}}{n} \right)^{n}} \bullet \frac{{2 + \frac{{3}}{n}}}{\left( 1 + \frac{{1}}{n} \right)^{3}}} = \frac{2}{e}$