An algebra problem by Vladimir Smith

$a_{n-1}=\frac{n}{n-2}(a_1+a_2+a_3+...+a_{n-2})$ From which it follows that: $a_1+a_2+a_3+...+a_{n-2}=\frac{n-2}{n}\cdot(a_{n-1})$ $a_n=\frac{n+1}{n-1}( a_1+a_2+a_3+...+a_{n-2}+a_{n-1})$ $a_n=\frac{n+1}{n-1}[\frac{n-2}{n}\cdot(a_{n-1})+a_{n-1}]$ $a_n=\frac{2(n+1)}{n}\cdot a_{n-1}$

The previous equation gives a telescoping product for $a_n$ such that $a_{2015}=\frac{2^{2014}\cdot2016!}{2\cdot2015!}=2016\cdot2^{2013}=63\cdot2^{2018}$ Finally, $b+c+d=63+2+2018=2083$ .

Let

$S_n=a_1+a_2+...+a_n$

$a_n=S_n-S_{n-1}$

Then we can rewrite equation as:

$S_n-S_{n-1}=\frac{n+1}{n-1} S_{n-1}$

$S_n=\frac{2n}{n-1} S_{n-1}$

Now you can easily see that:

$S_n=\frac{2^i n}{n-i} S_{n-i}$

Putting $i=1$ and $S_1=1$ you can calculate:

$S_{2015}=2^{2014} 2015$ and $S_{2014}=2^{2013} 2014$

And you will get:

$a_n=63 \dot{} 2^{2018}$