Solving Many Equations At Once

Let $A=\frac {x_{1}}{x_{1}+1}=\frac {x_{2}}{x_{2}+3}=...=\frac {x_{1007}}{x_{1007}+2013}=\frac {x_{1}+x_{2}+x_{3}+...+x_{1007}}{x_{1}+x_{2}+x_{3}+...+x_{1007}+1+3+5+...+2013}$ .

You can find that $1+3+5...+2013=(1007)^{2}$

and $x_{1}+x_{2}+x_{3}+...+x_{1007}=(2014)^{2}=4(1007)^{2}$ .

$\Rightarrow A=\frac {4(1007)^{2}}{4(1007)^{2}+(1007)^{2}}=\frac {4}{5}$ .

Since $A=\frac{x_{253}}{x_{253}+505}=\frac {4}{5}$

$\Rightarrow x_{253}=\boxed {2020}$ .

Let $a = \frac{x_{k}}{x_{k} + 2k - 1}$ for $1 \le k \le 1007$ .

Solving for $x_{k}$ , we get that $x_{k} = (2k-1) \frac{a}{1-a}$

Therefore, $2014^{2} = \sum_{k=1}^{1007} x_{k} = \frac{a}{1-a} \sum_{k=1}^{1007} (2k-1)$

Now the sum of the first 1007 odd numbers is just $1007^{2}$ , so we have: $2014^{2} = \frac{a}{1-a} * 1007^{2}$

Thus $4 = \frac{a}{1-a}$

Hence, from the formula above, we have: $x_{k} = (2k-1) \frac{a}{1-a} = 8k - 4$

Therefore, $x_{253} = 2020$

The simplest way to solve it is to inverse all the fractions from the second relation.

Then subtract one. And you will get.

$\frac {1}{{x}_{1}}=\frac {3}{{x}_{2}}=...=\frac {2k-1}{{x}_{k}}$

Then equate it with $\frac {1}{n}$ and you'll get that ${x}_{k} = n(2k - 1)$

The rest is a first degree equation we we plug in the first equation.

Invert all the equal ratios given in (2) and subtract 1 from them, which gives $x_{2} = 3 x_{1}$ ; $x_{3} = 5 x_{1}; . . . x_{n} = (2n-1) x_{1}$

So the sum in (1) can be written as $x_{1}\left(1+3+5+...+2013\right) = x_{1}(1007)^2 = 2014^2$ Hence $x_{1} = 4$ and $x_{253} = 4 \times (253 \times 2 - 1) = \boxed{2020}$

we can see, x1+x2+....+x1007=(2014)^2 or, x1+x2+....+x1007=(1007 * 2)^2 in this way we can write, x1+x2+....+x253=(253 * 2)^2 = 256036 in same way, x1+x2+....+x252=(252 * 2)^2 = 254016 now, x253 = (x1+x2+....+x253) - (x1+x2+....+x252) =256036 - 254016 =2020 in this solution u don't need the other equation

we can find all values of x2 and other terms upto x1007 in terms of x1 by equating and getting it as 3x1,5x1,......,2013x1 (Xk=(2k-1)x1) .substituting in 1st eqn we get x1 (1007^2)=2014^2. therfore x1=4. from the value of x1 we know that x253=(505 * x1)=505 4=2020

We note that

$\begin{aligned} \frac {x_1}{x_1+1} & = \frac {x_n}{x_n+(2n-1)} & \small \color{#3D99F6} \text{where } n \in \mathbb N \le 1007 \\ x_1x_n + (2n-1)x_1 & = x_1x_n+x_n \\ \implies x_n & = (2n-1)x_1 \end{aligned}$

Therefore, we have:

$\begin{aligned} x_1 + x_2+x_3 + \cdots + x_{1007} & = 2014^2 \\ x_1(1+3+5+\cdots+2013) & = 2014^2 \\ \frac {1007(1+2013)}2x_1 & = 2014^2 \\ 1007^2 x_1 & = 2014^2 \\ \implies x_1 & = 4\end{aligned}$

And that $x_{253} = (2\cdot 253-1)x_1 = 505 \cdot 4 = \boxed{2020}$ .

x1/(x1 +1) = x2/(x2 +3)

or 1/( 1+1/x1) = 1/(1 + 3/x2)

or x2 = 3*x1

Similarly we ca prove x3 = 5 x1, x4 = 7 x1 ........., x1007 = 2013*x1 so x1 + x2 + x3 + x4 +.................. + x1007

= x1 + 3 x1 + 5 x1 + 7*x1 + ................................ + 2013 * x1 ( Arith. series)

= x1 * 1007 * ( 1 + 2013) / 2 which is = 2014 * 2014 (given)

implies x1 = 4

Because xn = ( 2*n - 1 ) * x1

so x253 = ( 2 * 253 - 1) * 4 = 2020