One for All; All for One

Note first that $g(n) = n^{2}g(n) - (n - 1)^{2}g(n - 1)$ , and thus

$g(n) = \dfrac{(n - 1)^{2}}{n^{2} - 1}g(n - 1) = \dfrac{n - 1}{n + 1}g(n - 1)$ for $n \ge 2.$

Now observe that $g(2) = \dfrac{2}{2*3}, g(3) = \dfrac{2}{3*4}, g(4) = \dfrac{2}{4*5}, g(5) = \dfrac{2}{5*6}, \cdots.$

So it would appear that the general formula is $g(n) = \dfrac{2}{n(n + 1)}$ , which also holds for $g(1) = 1.$ Using induction, we have just observed that the formula hold for $n = 1,$ so now assume that it holds for some $n \ge 1.$ Then

$g(n + 1) = \dfrac{n}{n + 2}g(n) = \dfrac{n}{n + 2}*\dfrac{2}{n(n + 1)} = \dfrac{2}{(n + 1)(n + 2)}$ ,

and so the formula holds true for $n + 1$ given that it holds true for $n.$

Therefore $2015*g(2015) = 2015*\dfrac{2}{2015*2016} = \dfrac{1}{1008},$ and thus $a + b = 1 + 1008 = \boxed{1009}.$

It's see easy to see that $g(1) = 1, g(2) = \frac {1}{3} , g(3) = \frac 1 6 , g(4) = \frac {1}{10}, g(5) = \frac {1}{15}$ , which are all reciprocal of triangular numbers. So $g(n) = \frac {2}{n(n+1)}$ , I leave the proof for the reader, apply telescoping sum, it will satisfy the summation given. So the fraction is simply $2015 \times \frac {2}{2015 \times 2016} = \frac {1}{1008}$

You would need a pen and paper to properly understand the solution.......g(1) + g(2) + .........g(n-1) = (n-1)^2 g(n-1) = (n^2-1) g(n). ; (n-1) g( n-1) = ( n+1) *g( n). ; on writing this equation sequentially one by one by putting n=2, then n=3 and so on......then adding LHS and RHS of those equations respectively, we get...........g(1) = g(2) + g(3) + g(4) + ........g(n-1) + (n+1) g(n)......adding g(1) to both sides......and then using the first relation......we get......2 g(1) = n g(n) + (n^2 ) g(n)..........putting n=2015......we get 2015 *g(2015) = 2/2016 = 1/1008..........