| For All Solvers
NF杯2024

NF杯2024(S) - 積分を使わない別解

ユーザー解説 by kzy33550336

　漸化式より

xn+2xn+1(xn+2−xn+1)=xn+1xn(xn+1−xn)=⋯=x2x1(x2−x1)=314k(314−k)x_{n+2}x_{n+1}(x_{n+2}-x_{n+1})=x_{n+1}x_n(x_{n+1}-x_n)=\cdots=x_2x_1(x_2-x_1)=314k(314-k)

特に xn+1−xn=314k(314−k)xn+1xn>0x_{n+1}-x_n=\dfrac{314k(314-k)}{x_{n+1}x_n}\gt0 より xn<xn+1x_{n}\lt x_{n+1} である．上式を n=1n=1 から n=mkn=m_k まで足し上げると，

314kmk(314−k)=∑n=1mkxn+2xn+1(xn+2−xn+1)=13∑n=1mk(xn+23−xn+13−(xn+2−xn+1)3)=13(xmk+23−x23)−13∑n=1mk(xn+2−xn+1)3\begin{aligned} 314km_k(314-k)&=\sum_{n=1}^{m_k}x_{n+2}x_{n+1}(x_{n+2}-x_{n+1})\\ &=\frac{1}{3}\sum_{n=1}^{m_k}\left(x_{n+2}^3-x_{n+1}^3-(x_{n+2}-x_{n+1})^3\right)\\ &=\frac{1}{3}(x_{m_k+2}^3-x_2^3)-\frac{1}{3}\sum_{n=1}^{m_k}(x_{n+2}-x_{n+1})^3\\ \end{aligned}

ここで 0<xn<xn+1<xn+20\lt x_n\lt x_{n+1}\lt x_{n+2} より

xn+2−xn+1xn+1−xn=xnxn+2<1∴xn+2−xn+1<xn+1−xn<⋯<x3−x2\begin{aligned} \frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_{n}}&=\frac{x_n}{x_{n+2}}\lt1\quad \therefore x_{n+2}-x_{n+1}\lt x_{n+1}-x_n\lt\cdots\lt x_3-x_2\\ \end{aligned}

であり，x2<x3x_2\lt x_3 より

x3−x2=x1x3(x2−x1)<x1x2(x2−x1)=k(314−k)314x_3-x_2=\frac{x_1}{x_3}(x_2-x_1)\lt \frac{x_1}{x_2}(x_2-x_1)=\frac{k(314-k)}{314}

であるので，

0<∑n=1mk(xn+2−xn+1)3<∑n=1mk(x3−x2)3=mk(x3−x2)3<kmk⋅(314−k)33143⋅k2→0(k→+0)0\lt\sum_{n=1}^{m_k}(x_{n+2}-x_{n+1})^3\lt \sum_{n=1}^{m_k}(x_3-x_2)^3=m_k(x_3-x_2)^3\lt km_k\cdot\frac{(314-k)^3}{314^3}\cdot k^2\to0\quad(k\to+0)

従って lim⁡k→+0∑n=1mk(xn+2−xn+1)3=0\displaystyle\lim_{k\to+0}\sum_{n=1}^{m_k}(x_{n+2}-x_{n+1})^3=0 を得る．また，mkm_k の定義から xmk≤2024<xmk+1<xmk+2x_{m_k}\leq 2024\lt x_{m_k+1}\lt x_{m_k+2} となるので，

0<xmk+2−2024≤xmk+2−xmk=(xmk+2−xmk+1)+(xmk+1−xmk)<k(314−k)314+k(314−k)314→0(k→+0)\begin{aligned} 0&\lt x_{m_k+2}-2024\leq x_{m_k+2}-x_{m_k}\\ &=(x_{m_k+2}-x_{m_k+1})+(x_{m_k+1}-x_{m_k})\\ &\lt\frac{k(314-k)}{314}+\frac{k(314-k)}{314}\to0\quad(k\to+0)\\ \end{aligned}

従って xmk+2→2024 (k→+0)x_{m_k+2}\to2024\ (k\to+0) となる．以上より，

lim⁡k→+0kmk=lim⁡k→+01314(314−k){13(xmk+23−x23)−13∑n=1mk(xn+2−xn+1)3}=1314(314−0){13(20243−x23)−13⋅0}=20243−31433⋅3142=68837589024649\begin{aligned} \lim_{k\to+0}km_k&=\lim_{k\to+0}\frac{1}{314(314-k)}\left\{\frac{1}{3}(x_{m_k+2}^3-x_2^3)-\frac{1}{3}\sum_{n=1}^{m_k}(x_{n+2}-x_{n+1})^3\right\}\\ &=\frac{1}{314(314-0)}\left\{\frac{1}{3}(2024^3-x_2^3)-\frac{1}{3}\cdot0\right\}\\ &=\frac{2024^3-314^3}{3\cdot314^2}=\frac{688375890}{24649} \end{aligned}