A nice sum

\hspace{-20}$Given $\bf{x_{1}+x_{2}+x_{3}+............+x_{1007}=(2014)^2..........(1)}$\\\\\\ And $\bf{\frac{x_{1}}{x_{1}+1}=\frac{x_{2}}{x_{2}+3}=\frac{x_{3}}{x_{3}+5}=...........=\frac{x_{1007}}{x_{1007}+2013}}$\\\\\\ Now Taking Reciprocal ..........\\\\\\ $\bf{\Rightarrow \frac{x_{1}+1}{x_{1}}=\frac{x_{2}+3}{x_{2}}=\frac{x_{3}+5}{x_{3}}=...........=\frac{x_{1007}+2013}{x_{1007}}}$\\\\\\ Now Sub. $\bf{1}$ from each Term, we get......\\\\\\ $\bf{\Rightarrow \frac{1}{x_{1}}=\frac{3}{x_{2}}=\frac{5}{x_{3}}=.............=\frac{2013}{x_{1007}}}$\\\\\\ Now again Taking Reciprocal......\\\\\\ Let $\bf{\frac{x_{1}}{1}=\frac{x_{2}}{3}=\frac{x_{3}}{5}=............=\frac{x_{1007}}{2013}=k.............(2)}$\\\\\\ So $\bf{x_{1}+x_{2}+x_{3}+..........+x_{1007}=k\cdot \left(1+3+5+......+1007\right)=(1007)^2}$\\\\\\ Now from equation......$\bf{(1)}$\\\\\\ $\bf{\Rightarrow (2014)^2=(1007)\Rightarrow k = 4}$\\\\\\ So from equation..........\bf{(2)} \\\\\\

\hspace{-20}\bf{\frac{x_{1}}{1}=\frac{x_{2}}{3}=\frac{x_{3}}{5}=.....=\frac{x_{n}}{(2n-1)}.......=\frac{x_{1007}}{2013}=k}$\\\\\\ So $\bf{\frac{x_{n}}{(2n-1)}=k\Rightarrow x_{n}=k\cdot (2n-1)}$\\\\\\ Now Put $\bf{n=253\;,k=4\;,}$ we get\\\\\\ $\bf{\Rightarrow x_{253}=4\cdot (2\times 253-1)=4\cdot 505=2020}$

The general term is x_r/(x_r +2r-1).Yours is also right.

Is it 2020?