자연계열 B 3번

(3-1) $\displaystyle \frac{a_n}{2^n}=\int_0^1x^n(1-x)^{102-n}dx$이므로$\displaystyle \begin{aligned}\frac{a_{98}}{2^{98}}+3\frac{a_{99}}{2^{99}}+3\frac{a_{100}}{2^{100}}+\frac{a_{101}}{2^{101}}&=\int_0^1(x^{98}(1-x)^4+3x^{99}(1-x)^3+3x^{100}(1-x)^2+x^{101}(1-x))dx\\&=\int_0^1x^{98}(1-x)((1-x)^3+3x(1-x)^2+3x^2(1-x)+x^3)dx\\&=\int_0^1x^{98}(1-x)((1-x)+x)^3dx=\int_0^1(x^{98}-x^{99})dx\\&=\frac1{99}-\frac1{100}=\frac1{9900}\end{aligned}$그러므로 $\displaystyle p=9900$, $\displaystyle q=1$이다.

(3-2) $\displaystyle \begin{aligned}a_n&=2^n\int_0^1x^n(1-x)^{102-n}dx=2^n\Biggl[\frac{x^{n+1}}{n+1}(1-x)^{102-n}\Biggr]_0^1+2^n\int_0^1\frac{x^{n+1}}{n+1}(102-n)(1-x)^{101-n}dx\\&=2^n\frac{102-n}{n+1}\int_0^1x^{n+1}(1-x)^{101-n}dx=\frac{102-n}{2(n+1)}a_{n+1}\end{aligned}$이므로 $\displaystyle \frac{a_n}{a_{n+1}}=\frac{102-n}{2(n+1)}$이다.

$\displaystyle \frac{a_n}{a_{n+1}}=\frac{102-n}{2(n+1)}\ge1$이 되려면 $\displaystyle 102-n\ge2n+2$이므로 $\displaystyle n\le\frac{100}3$이다. 그러므로 $\displaystyle n=1,2,3,\ldots,33$이다.

(3-3) $\displaystyle a_1>a_2>\cdots>a_{33}>a_{34}<a_{35}<\cdots<a_{100}<a_{101}$이므로 $\displaystyle m=34$이다.

그런데 $\displaystyle {}_{102}\mathrm C_{34}\times a_{34}={}_{102}\mathrm C_{34}\times\frac{a_{34}}{a_{35}}\times\frac{a_{35}}{a_{36}}\times\cdots\times\frac{a_{99}}{a_{100}}\times\frac{a_{100}}{a_{101}}\times a_{101}$이고$\displaystyle a_{101}=\int_0^1(2x)^{101}(1-x)dx=2^{101}\int_0^1(x^{101}-x^{102})dx=2^{101}\Biggl[\frac{x^{102}}{102}-\frac{x^{103}}{103}\Biggr]_0^1=\frac{2^{101}}{102\times103}$이므로$\displaystyle {}_{102}\mathrm C_{34}\times a_{34}={}_{102}\mathrm C_{34}\times\frac{68}{2\times35}\times\frac{67}{2\times36}\times\cdots\times\frac3{2\times100}\times\frac2{2\times101}\times\frac{2^{101}}{102\times103}=\frac{2^{34}}{103}$이다.