Đáp án: $\dfrac{2}{2017}$
Giải thích các bước giải:
Ta có:
$1+\dfrac{1}{n(n+2)}=\dfrac{n(n+2)+1}{n(n+2)}=\dfrac{n^2+2n+1}{n(n+2)}=\dfrac{(n+1)^2}{n(n+2)}$
Áp dụng
$\to A=\dfrac{(1+1)^2}{1(1+2)}\cdot \dfrac{(2+1)}{2(2+2)^2}\cdots \dfrac{(2015+1)^2}{2015(2015+2)}\cdot \dfrac{1}{2016}$
$\to A=\dfrac{2^2}{1.3}\cdot \dfrac{3^2}{2.4}\cdots \dfrac{2016^2}{2015.2017}\cdot \dfrac{1}{2016}$
$\to A=\dfrac{2.3..2016}{1.2...2015}.\dfrac{2.3...2016}{3.4...2017}.\dfrac{1}{2016}$
$\to A=2016.\dfrac{2}{2017}.\dfrac1{2016}$
$\to A=\dfrac{2}{2017}$
