Queen Propertyの遊び場
Queen Propertyとは
Queen Propertyという名前は
すい
さんが積分におけるKing Propertyに擬えて名付けた和を求めるテクニックの名称です。
積分においては、$$ \int_a^b f(x)dx=\dfrac12 \int_a^b \{f(x)+f(a+b-x)\}dx$$が成り立ち、これがKing Propertyと呼ばれるものです。
一方でただの和についても、$$ \displaystyle\sum_{k=a}^b f(k)=\dfrac12 \displaystyle\sum_{k=a}^b \{f(k)+f(a+b-k)\}$$が成り立ち、これをQueen Propertyと呼んでいます。この式は証明するまでもなく、逆順から足し合わせた$$ \displaystyle\sum_{k=a}^b f(a+b-k)$$が元の和に等しいことから、ほぼ自明です。しかしながら、そんな「あたりまえ」の事実から、面白い問題が生まれるというのが興味深いです。
以下では、種々のQueen Propertyを用いた問題とその鮮やかな解法を紹介します。
問題集
普通型
$$\displaystyle\sum_{k=0}^n\dfrac{k^3}{k^3+(n-k)^3}$$
解答
\begin{eqnarray} \displaystyle\sum_{k=0}^n\dfrac{k^3}{k^3+(n-k)^3} &=& \dfrac12 \displaystyle\sum_{k=0}^n\left\{\dfrac{k^3}{k^3+(n-k)^3}+\dfrac{(n-k)^3}{(n-k)^3+k^3}\right\} \\ &=& \dfrac12\displaystyle\sum_{k=0}^n1\\&=&\mathbf{\dfrac{n+1}{2}} \end{eqnarray}
$$\displaystyle\sum_{k=0}^n\dfrac{2k^2+n^2}{k^2-nk+n^2}$$
解答
\begin{align} \displaystyle\sum_{k=0}^n\dfrac{2k^2+n^2}{k^2-nk+n^2} &= \dfrac12 \displaystyle\sum_{k=0}^n\left\{\dfrac{2k^2+n^2}{k^2-nk+n^2}+\dfrac{2(n-k)^2+n^2}{(n-k)^2-n(n-k)+n^2}\right\} \\ &=\dfrac12\displaystyle\sum_{k=0}^n4 \\ &= \mathbf{2n+2} \end{align}
三角関数型
$$\displaystyle\sum_{n=0}^{90}\dfrac{(\sin n^\circ +2)(4\sin n^\circ -1)}{\sin n^\circ+\cos n^\circ}$$
解答
\begin{align} \displaystyle\sum_{n=0}^{90}\dfrac{(\sin n^\circ +2)(4\sin n^\circ -1)}{\sin n^\circ+\cos n^\circ} &= \dfrac12 \displaystyle\sum_{n=0}^{90}\dfrac{(\sin n^\circ +2)(4\sin n^\circ -1)+(\cos n^\circ +2)(4\cos n^\circ -1)}{\sin n^\circ+\cos n^\circ} \\ &=\dfrac12 \displaystyle\sum_{n=0}^{90}\dfrac{(4\sin^2 n^\circ+7\sin n^\circ-2)+(4\cos^2 n^\circ+7\cos n^\circ-2)}{\cos n^\circ+\cos n^\circ}\\ &= \dfrac12 \displaystyle\sum_{n=0}^{90}7 \\&= \mathbf{\dfrac{637}{2}}\end{align}
$$\displaystyle\sum_{k=1}^{89}\dfrac{1}{1+\tan^3(k^\circ)}$$
解答
\begin{align} \displaystyle\sum_{n=1}^{89}\dfrac{1}{1+\tan^3(k^\circ)} &= \dfrac12 \displaystyle\sum_{n=1}^{89} \left\{\dfrac{1}{1+\tan^3(k^\circ)}+\dfrac{1}{1+\dfrac{1}{\tan^3(k^\circ)}}\right\} \\ &=\dfrac12 \displaystyle\sum_{n=1}^{89}1\\ &=\mathbf{\dfrac{89}{2}}\end{align}
$$\displaystyle\sum_{k=1}^{179}\log \sin^k(k^\circ)$$
解答
\begin{align} \displaystyle\sum_{k=1}^{179}\log \sin^k(k^\circ) &=\displaystyle\sum_{k=1}^{179}k\log \sin(k^\circ)\\ &=\dfrac12 \displaystyle\sum_{k=1}^{179}\left\{k\log \sin(k^\circ)+(180-k)\log \sin((180-k)^\circ)\right\}\\&= 90\displaystyle\sum_{k=1}^{179}\log \sin(k^\circ)\\&=90\log\prod_{k=1}^{179}\sin(k^\circ)\\ &= 180\log \prod_{k=1}^{89}\sin(k^\circ)\\ &=180\log\sqrt{\dfrac{180}{2^{179}}}\\&=90(\log180-179\log2)\\ &=90(2\log3+\log5-177\log2) \end{align}
指数関数型
$$\displaystyle\sum_{k=1}^{99}\dfrac{4^\dfrac{k}{100}}{4^\dfrac{k}{100}+2}$$
解答
\begin{align} \displaystyle\sum_{k=1}^{99}\dfrac{4^\dfrac{k}{100}}{4^\dfrac{k}{100}+2} &=\dfrac12 \displaystyle\sum_{k=1}^{99}\left(\dfrac{4^\dfrac{k}{100}}{4^\dfrac{k}{100}+2}+\dfrac{4^{1-\dfrac{k}{100}}}{4^{1-\dfrac{k}{100}}+2}\right)\\ &=\dfrac12 \displaystyle\sum_{k=1}^{99}\left(\dfrac{4^\dfrac{k}{100}}{4^\dfrac{k}{100}+2}+\dfrac{4}{4+2\cdot4^{\dfrac{k}{100}}}\right)\\&= \dfrac12 \displaystyle\sum_{k=1}^{99}\left(\dfrac{4^\dfrac{k}{100}}{4^\dfrac{k}{100}+2}+\dfrac{2}{2+4^{\dfrac{k}{100}}}\right)\\&=\mathbf{\dfrac{99}2}\end{align}
$$\displaystyle\sum_{k=0}^{2n}\dfrac{1}{2^n+2^k}$$
解答
\begin{align} \displaystyle\sum_{k=0}^{2n}\dfrac{1}{2^n+2^k} &=\dfrac12 \displaystyle\sum_{k=0}^{2n}\left(\dfrac{1}{2^n+2^k}+\dfrac{1}{2^n+2^{2n-k}}\right)\\ &= \dfrac12 \displaystyle\sum_{k=0}^{2n}\left(\dfrac{1}{2^n+2^k}+\dfrac{2^{k-n}}{2^k+2^n}\right)\\&= \dfrac1{2^{n+1}} \displaystyle\sum_{k=0}^{2n}\left(\dfrac{2^n}{2^n+2^k}+\dfrac{2^k}{2^k+2^n}\right)\\&=\mathbf{\dfrac{2n+1}{2^{n+1}}}\end{align}
二項係数型
$$\displaystyle\sum_{k=0}^n k\binom nk$$
解答
\begin{align} \displaystyle\sum_{k=0}^n k\binom nk &=\dfrac12 \displaystyle\sum_{k=0}^n \left(k\binom nk +(n-k)\binom{n}{n-k}\right) \\ &= \dfrac12 \displaystyle\sum_{k=0}^n n\binom nk \\&= \mathbf {n2^{n-1}}\end{align}
$$\displaystyle\sum_{k=0}^n \dfrac{\binom n{k-1}^2}{\binom n{k-1}-\binom n{k+1}}$$(nは奇数)
解答
\begin{align} \displaystyle\sum_{k=1}^{n-1} \dfrac{\binom n{k-1}^2}{\binom n{k-1}-\binom n{k+1}} &= \dfrac12 \displaystyle\sum_{k=1}^{n-1} \left(\dfrac{\binom n{k-1}^2}{\binom n{k-1}-\binom n{k+1}}+\dfrac{\binom n{n-k-1}^2}{\binom n{n-k-1}-\binom n{n-k+1}}\right) \\&= \dfrac12 \displaystyle\sum_{k=1}^{n-1} \left(\dfrac{\binom n{k-1}^2}{\binom n{k-1}-\binom n{k+1}}+\dfrac{\binom n{n-k-1}^2}{\binom n{n-k-1}-\binom n{n-k+1}}\right) \\&= \dfrac12 \displaystyle\sum_{k=1}^{n-1} \left(\dfrac{\binom n{k-1}^2}{\binom n{k-1}-\binom n{k+1}}+\dfrac{\binom n{k+1}^2}{\binom n{k+1}-\binom n{k-1}}\right) \\ &= \dfrac12 \displaystyle\sum_{k=1}^{n-1} \dfrac{\binom n{k-1}^2-\binom n{k+1}^2}{\binom n{k-1}-\binom n{k+1}} \\&= \dfrac12 \displaystyle\sum_{k=1}^{n-1}\left( \binom{n}{k-1}+\binom{n}{k+1} \right) \\&= \dfrac12(2^n-n-1+2^n-n-1) \\&= \mathbf{2^n-n-1} \end{align}
特殊型
$$\displaystyle\sum_{k=1}^{100}\left\lfloor \dfrac{k^3}{100} \right\rfloor$$
解答
\begin{align} \displaystyle\sum_{k=1}^{100}\left\lfloor \dfrac{k^3}{100} \right\rfloor &= \dfrac12 \displaystyle\sum_{k=0}^{100}\left(\left\lfloor \dfrac{k^3}{100} \right\rfloor+\left\lfloor \dfrac{{(100-k)}^3}{100} \right\rfloor\right) \\&= \dfrac12 \displaystyle\sum_{k=0}^{100}\left(\left\lfloor \dfrac{k^3}{100} \right\rfloor+\left\lfloor \dfrac{100^3-3\cdot100^2k+300k^2-k^3}{100} \right\rfloor\right) \\&= \dfrac12 \displaystyle\sum_{k=0}^{100}\left(\left\lfloor \dfrac{k^3}{100} \right\rfloor+\left\lfloor \dfrac{-k^3}{100} \right\rfloor\right)+255025 \end{align}
ここで、$$\left\lfloor \dfrac{k^3}{100} \right\rfloor+\left\lfloor \dfrac{-k^3}{100} \right\rfloor$$
は、$k$が$10$の倍数であれば$0,$そうでなければ$-1$となるので、元の和は、$$ \dfrac12(-100+10)+255025=255025-45=254980$$
関数$A(n),B(n)$を次により定める。
$A(n)=(0\le x \le n$を満たす$1001$と互いに素な整数$x$の個数$)$
$B(n)=(n\le x \le 1001$を満たす$1001$と互いに素な整数$x$の個数$)$
$$ \displaystyle\sum_{n=1}^{1000}\dfrac{A(n)^2}{A(n)-B(n)}$$を求めよ。ただし、任意の$1\le k\le1000$なる整数$k$に対して、$A(k)\ne B(k)$を保証する。
解答
$1001=7 \cdot 11 \cdot 13$であるから, $\phi(1001)=720$である.
ここで,ユークリッドの互除法を用いることで,
$$ \gcd(1001,1001-n)=\gcd(1001,n) $$
が成り立つことがわかる.特に,$A(n)$=$B(1001-n)$が成り立つ.
すると,
\begin{align} \sum_{n=1}^{1000} \frac{A(n)^2}{A(n)-B(n)}&=\left(\frac 12\sum_{n=1}^{1000} \frac{A(n)^2}{A(n)-B(n)}+ \frac{B(n)^2}{B(n)-A(n)} \right) \\&=\frac 12\sum_{n=1}^{1000} \frac{A(n)^2-B(n)^2}{A(n)-B(n)}\\&= \frac 12\sum_{n=1}^{1000} {\left(A(n)+B(n)\right)} \end{align}
ここで,
$$ A(n)+B(n) = \begin{cases} 721 & (\gcd(1001,n)=1) \\ 720 & (otherwise) \end{cases} $$
である.よって問題の総和は$$ \frac 12 \cdot720 \cdot 721+280 \cdot 720 = \frac 12 720720 = \textbf{360360}$$
終わり
At just seven years old, Gauss stunned his teacher by instantly calculating the sum of all integers from 1 to 100.His ingenious pairing—1+100, 2+99, and so on—revealed a mathematical mind far beyond his years.
