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Solve the system of $n$ equations and $n$ unknowns which the equations are $\left (L_{i} \right ): x_{i}-x_{i+1}=1$ for $i=1,2,....,n-1$ and $\left (L_{n} \right ):x_{n}=1$

$\left\{\begin{matrix} x_{i} - x_{i+1}=1\\ x_{n}=1 \end{matrix}\right.$

$\left\{\begin{matrix} x_{1} - x_{2}=1\\ x_{n}=1 \end{matrix}\right.$

$\left\{\begin{matrix} x_{2} - x_{3}=1\\ x_{n}=1 \end{matrix}\right.$

$\vdots $

$\left\{\begin{matrix} x_{n-1} - x_{n}=1\\ x_{n}=1 \end{matrix}\right.$

$\left\{\begin{matrix} x_{n-1} =2\\ x_{n}=1 \end{matrix}\right.$

solution is $\left ( 2;1 \right )$

is it correct guys ? Thanks for looking

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    You are supposed to find a solution $\forall$ $x_i$. – 2017-02-26

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Then, we will also have $$x_{n-2}-x_{n-1} = 1\Rightarrow x_{n-2} = 3$$Similarly $$x_{n-3}-x_{n-2} =1 \Rightarrow x_{n-3}=4$$ $$\vdots$$ $$ x_1-x_2 =1\Rightarrow x_1 = n $$ as we are supposed to find a solution for all $i $.

Hope it helps.

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    Riiiiiiiiight how could i forget about it . Thanks man it helps a lot . – 2017-02-26
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    we can also say that the sum of solutions is $(n+1;n)$ and $n\in \mathbb{N}^{\ast }$ – 2017-02-26