An algebra problem by Ilham Saiful Fauzi

Call the expression $s_{2016}$ .

We prove by induction that $s_n=n$ for even $n$ .

When $n=2$ , $s_2=\frac21=2$

Assume $s_k=k$ holds true for some even $k$ .

$s_{k+2}=\frac{k+2}{k+1}+\frac{k+2}{k+1}s_k$ $=\frac{k+2}{k+1}(1+k)$ $=k+2$

Thus by the principle of mathematical induction, we have shown the $s_n=n$ or in particular $s_{2016}=\boxed{2016}$ .

${ a }_{ n }=\quad \frac { \prod _{ k=1 }^{ n }{ (2018-2k) } }{ \prod _{ k=1 }^{ n }{ (2017-2k) } } ,\quad \forall \quad n\ge 2\\ \\ \therefore \quad { a }_{ n }=\quad \frac { \prod _{ k=1 }^{ n }{ (2018-2k) } \times \left[ \left\{ 2017-2k \right\} -\left\{ 2016-2k \right\} \right] }{ \prod _{ k=1 }^{ n }{ (2017-2k) } } \\ =\quad \frac { \prod _{ k=1 }^{ n }{ (2018-2k) } }{ \prod _{ k=1 }^{ n-1 }{ (2017-2k) } } \quad -\quad \frac { \prod _{ k=1 }^{ n+1 }{ (2018-2k) } }{ \prod _{ k=1 }^{ n }{ (2017-2k) } } \quad \left[ F(x)-F(x-1)\quad form,\quad hence\quad Telescopic \right] \\ Now,\quad \sum _{ n=2 }^{ 1008 }{ { a }_{ n } } =\quad F(2)-F(1008)=\quad \frac { 2016\times 2014 }{ 2015 } \quad (\quad Telescopic\quad series\quad )\\ \therefore \quad S=\quad \sum _{ n=2 }^{ 1008 }{ { a }_{ n } } +\quad \frac { 2016 }{ 2015 } =\quad \boxed { 2016 } \\ \\ \\ \\$