Problem 2. Solve the equation:

$$
\frac{x^{2}+1}{2}+\frac{2 x^{2}+1}{3}+\frac{3 x^{2}+1}{4}+\cdots+\frac{2015 x^{2}+1}{2016}=2015
$$
Solution:

The given equation is equivalent to:

$\left(\frac{x^{2}+1}{2}-1\right)+\left(\frac{2 x^{2}+1}{3}-1\right)+\left(\frac{3 x^{2}+1}{4}-1\right)+\cdots+\left(\frac{2015 x^{2}+1}{2016}-1\right)=0$

$.2 p$

After performing the calculations in each parenthesis, we obtain:

$\frac{x^{2}-1}{2}+\frac{2 x^{2}-2}{3}+\frac{3 x^{2}-3}{4}+\cdots+\frac{2015 x^{2}-2015}{2016}=0$

$1 p$

We factor out the common factor in the numerators where applicable and get:

$\frac{x^{2}-1}{2}+\frac{2\left(x^{2}-1\right)}{3}+\frac{3\left(x^{2}-1\right)}{4}+\cdots+\frac{2015\left(x^{2}-1\right)}{2016}=0$

$1p$

We factor out $x^{2}-1$ and have:

$\left(x^{2}-1\right)\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{2015}{2016}\right)=0$

$1p$

But the second factor is a positive number, being a sum of strictly positive numbers. $.1 p$

Thus $x^{2}-1=0 \Rightarrow x= \pm 1$

We obtain: $\quad S=\{1 ;-1\}$ $1p$