P = \(1+\dfrac{1}{2}\cdot\left(1+2\right)+\dfrac{1}{3}\cdot\left(1+2+3\right)+\dfrac{1}{4}\cdot\left(1+2+3+4\right)+...+\dfrac{1}{16}\cdot\left(1+2+3+...+16\right)\)

Cho a + b + c = 2016 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}.\) Tính S = \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

Giải giúp mình nha.