Question

For an arbitrary positive integer$\mathit{n}\mathbf{\ge }\mathbf{3}$, find all solutions ${\mathit{x}}_{\mathbf{1}}\mathbf{,}{\mathit{x}}_{\mathbf{2}}\mathbf{,}{\mathit{x}}_{\mathbf{3}}\mathbf{,}\mathbf{.}\mathbf{.....}\mathbf{,}{\mathit{x}}_{\mathbf{n}}$of the simultaneous equations ${\mathit{x}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{3}}\mathbf{)}\mathbf{,}{\mathit{x}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{4}}\mathbf{)}\mathbf{,}\mathbf{.}\mathbf{....}\mathbf{,}{\mathit{x}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{x}}_{\mathbf{n}\mathbf{-}\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{n}}\mathbf{)}$. Note that we are asked to solve the simultaneous equations ${\mathit{x}}_{\mathbf{k}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{x}}_{\mathbf{k}\mathbf{-}\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{)}$, for$\mathit{k}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{.}\mathbf{....}\mathbf{,}\mathit{n}\mathbf{-}\mathbf{1}$ .

Short answer

$x_n=3x_{n+2}-2x_{n+3}$is the expression using which the solutions$x_1,x_2,x_3,\mathrm{......},x_n$of the simultaneous equations can be obtained.

Step-by-step solution

Consider the given equations.

The expression for$x_2$is,

$x_2=\frac{1}{2}(x_1+x_3)$

Substitute the above equation in$x_3$

$\begin{array}{l}{x}_3=\frac{1}{2}(x_2+x_4)\\ \Rightarrow x_3=\frac{1}{2}(\left(\frac{1}{2}(x_1+x_3)\right)+x_4)\\ \Rightarrow x_3=\frac{1}{4}{x}_1+\frac{1}{2}{x}_3+\frac{1}{2}{x}_4\\ \therefore x_1=3x_3-2x_4\end{array}$

Compute the expressions for remaining variables.

Similarly,

$x_2=3x_4-2x_5$

Continuing the same, the expression for the $n^{th}$ term will be,

$x_n=3x_{n+2}-2x_{n+3}$

Compute the expressions for remaining variables.

Similarly,

$x_2=3x_4-2x_5$

Continuing the same, the expression for the $n^{th}$ term will be,

$x_n=3x_{n+2}-2x_{n+3}$

The solutions $x_1,x_2,x_3,\mathrm{......},x_n$of the simultaneous equations can be found using the expression,$x_n=3x_{n+2}-2x_{n+3}$ .