An algebra problem by Aira Thalca

The value of $f(2)$ is

$f(1) + f(2) = 4\times f(2)$

$2016 + f(2) = 4\times f(2)$

$3\times f(2) = 2016$

$f(2) = \frac{2016}{3} = \frac{2016}{1+2}$

The value of $f(3)$ is

$f(1) + f(2) + f(3) = 9\times f(3)$

$2016 + f(2) + f(3) = 9\times f(3)$

$8\times f(2) = 2016 + \frac{2016}{3}$

$f(3) = \frac{2016}{6} = \frac{2016}{1+2+3}$

So,

$f(x) = \frac{2016}{1 + 2 + ... + n}$

Then given

$f(2016) = \frac{2016\times 2}{2016\times 2017}$

$f(2016) = \frac{2}{2017}$

Then given

$a + b = 2 + 2017 = 2019$

$\color{#3D99F6}{f(1) + f(2) + \cdots + f(n-1) = (n-1)^{2} \times f(n-1)}$

$\begin{aligned} \color{#3D99F6}{f(1) + f(2) + \cdots + f(n-1)} \color{#333333}{+ f(n)} & = n^2 \times f(n) \\ \color{#3D99F6}{(n-1)^{2} \times f(n-1)} \color{#333333}{+ f(n)} & = n^2 \times f(n) \\ (n-1)^{2} \times f(n-1)& = (n^2-1) \times f(n) \\ (n-1) \times f(n-1)& = (n+1) \times f(n) \\ f(n)&=\dfrac{n-1}{n+1} \times f(n-1) \\ f(n)&=\dfrac{\cancel{n-1}}{n+1} \times \dfrac{\cancel{n-2}}{n} \times \dfrac{\cancel{n-3}}{\cancel{n-1}} \times \cdots \times \dfrac{ \cancel{3}}{\cancel{5}} \times \dfrac{2}{\cancel{4}} \times \dfrac{1}{\cancel{3}} \times 2016\\ f(n)&=\dfrac{2}{n(n+1)} \times 2016 \end{aligned}$

Putting $n=2017$

$\begin{aligned} f(2016)&=\dfrac{2}{2016(2016+1)} \times 2016 \\ f(2016)&=\dfrac{2}{2017} \end{aligned}$

$a=2 \text{, } b=2017 \text{. }a+b=2+2017=\boxed{19}$