题解：P5389 [Cnoi2019] 数学课
Nygglatho
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题解
比较目光呆滞的推柿子：
\begin{aligned}
P(x=y)&=\sum\limits_{v_1,v_2}\dfrac{\min(v_1v_2)}{v_1v_2}\\
&=\sum\limits_{i=1}^n\sum\limits_{j=1}^n\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{\min(a_i,a_j)}{a_ia_j}\\
&=\sum\limits_{i=1}^n\sum\limits_{j=1}^i\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{1}{a_i}+\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{1}{a_j}\\
&=\sum\limits_{i=1}^n\sum\limits_{j=1}^i\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{2}{i(i+1)}+\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{2}{j(j+1)}\\
&=\sum\limits_{i=1}^n\sum\limits_{j=1}^i\dfrac{6}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}+\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{6}{n(n+1)(n+2)}\\
&=\sum\limits_{i=1}^n\sum\limits_{j=1}^i\dfrac{6}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}+\sum\limits_{i=1}^n\dfrac{3i(i+1)(n-i)}{n(n+1)(n+2)}\dfrac{6}{n(n+1)(n+2)}\\
&=\dfrac{18}{n^2(n+1)^2(n+2)^2}\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^ij(j+1)+\sum\limits_{i=1}^ni(i+1)(n-i)\right)\\
&=\dfrac{18}{n^2(n+1)^2(n+2)^2}\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^i\left(j^2+j\right)+\sum\limits_{i=1}^n\left(-i^3+(n-1)i^2+ni\right)\right)\\
&=\dfrac{18}{n^2(n+1)^2(n+2)^2}\left(\sum\limits_{i=1}^n\left(\dfrac{i(i+1)(i+2)}{3}\right)+\sum\limits_{i=1}^n\left(-i^3+(n-1)i^2+ni\right)\right)\\
&=\dfrac{18}{n^2(n+1)^2(n+2)^2}\left(\dfrac 1 3\sum\limits_{i=1}^n\left(i^3+3i^2+2i\right)+\sum\limits_{i=1}^n\left(-i^3+(n-1)i^2+ni\right)\right)\\
&=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{n^2(n+1)^2+2n(n+1)(2n+1)+4n(n+1)-3n^2(n+1)^2+2n(n-1)(n+1)(2n+1)+6n^2(n+1)}{12}\\
&=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{-2n^2(n+1)^2+2n^2(n+1)(2n+1)+4n(n+1)+6n^2(n+1)}{12}\\
&=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{n(n+1)-2n(n+1)+2n(2n+1)+6n+4}{12}\\
&=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{n(n+1)\left(2n^2+6n+4\right)}{12}\\
&=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{n(n+1)^2(n+2)}{6}\\
&=\dfrac{3}{n(n+2)}\\
\end{aligned}