자연계열 1-2번

임의의 실수 $\displaystyle a, b$에 대하여 $\displaystyle f(a+b)=f(a)+f(b)$이므로,$\displaystyle f(0)=f(0+0)=f(0)+f(0)$이고, $\displaystyle 6=f(3)=f(2+1)=f(2)+f(1)=f(1+1)+f(1)=f(1)+f(1)+f(1)$.따라서 $\displaystyle f(0)=0, f(1)=2, f(2)=4$. 또한, $\displaystyle f(2-x)=f(2)-f(x)$이므로$\displaystyle f(x)=t$로 치환해서 적분하면$\displaystyle {}_{20}\mathrm{C}_{10}\int_{0}^{2}f(x)^{9}f(2-x)^{10}f^{\prime}(x)\,dx={}_{20}\mathrm{C}_{10}\int_{0}^{4}t^{9}(4-t)^{10}\,dt$이고,

$\displaystyle I=\int_{0}^{4}t^{9}(4-t)^{10}\,dt$에 부분적분을 적용하면$\displaystyle \begin{aligned}I&=\int_{0}^{4}t^{9}(4-t)^{10}\,dt=\Biggl[\frac{1}{10}t^{10}(4-t)^{10}\Biggr]_{0}^{4}+\int_{0}^{4}t^{10}(4-t)^{9}\,dt\\&=\int_{0}^{4}t^{10}(4-t)^{9}\,dt=\Biggl[\frac{1}{11}t^{11}(4-t)^{9}\Biggr]_{0}^{4}+\frac{9}{11}\int_{0}^{4}t^{11}(4-t)^{8}\,dt\\&=\frac{9}{11}\int_{0}^{4}t^{11}(4-t)^{8}\,dt=\frac{9}{11}\frac{8}{12}\int_{0}^{4}t^{12}(4-t)^{7}\,dt\\&=\cdots=\frac{9}{11}\frac{8}{12}\frac{7}{13}\cdots\frac{1}{19}\int_{0}^{4}t^{19}(4-t)^{0}\,dt\\&=\frac{9}{11}\frac{8}{12}\frac{7}{13}\cdots\frac{1}{19}\frac{4^{20}}{20}\end{aligned}$따라서$\displaystyle {}_{20}\mathrm{C}_{10}I={}_{20}\mathrm{C}_{10}\frac{9}{11}\frac{8}{12}\cdots\frac{1}{19}\frac{4^{20}}{20}=\frac{2^{40}}{10}=\frac{2^{39}}{5}.$