Giải thích các bước giải:
Ta có :
$A^2_n=\dfrac{n!}{(n-2)!}=n(n-1)\to\dfrac{1}{A^2_n}=\dfrac{1}{n(n-1)}=\dfrac{1}{n-1}-\dfrac{1}{n} $
$\to \dfrac{1}{A^2_2}+\dfrac{1}{A^2_3}+\dfrac{1}{A^2_4}+..+\dfrac{1}{A^2_n}=\dfrac{2015}{2016}$
$\to \dfrac 11-\dfrac 12+\dfrac 12-\dfrac 23+..+\dfrac{1}{n-1}-\dfrac{1}{n}=\dfrac{2015}{2016}$
$\to 1-\dfrac{1}{n}=\dfrac{2015}{2016}$
$\to \dfrac 1n=1-\dfrac{2015}{2016}=\dfrac{1}{2016}$
$\to n=2016$