# Question #97138

Nov 10, 2016

Proposing a solution with the structure

${a}_{n} = C {\lambda}^{n}$ and substituting into the difference equation

${a}_{n + 2} - {a}_{n + 1} - {a}_{n} = 0$

we have

$C {\lambda}^{n + 2} - C {\lambda}^{n + 1} - C {\lambda}^{n} = C {\lambda}^{n} \left({\lambda}^{2} - \lambda - 1\right) = 0$ so the values

$\lambda = 0 , \lambda = \frac{1 \pm \sqrt{5}}{2}$ are eventual solutions for the difference equation.

So ${a}_{n} = {C}_{0} {0}^{n} + {C}_{1} {\left(\frac{1 + \sqrt{5}}{2}\right)}^{n} + {C}_{2} {\left(\frac{1 - \sqrt{5}}{2}\right)}^{n} = {C}_{1} {\left(\frac{1 + \sqrt{5}}{2}\right)}^{n} + {C}_{2} {\left(\frac{1 - \sqrt{5}}{2}\right)}^{n}$

Now, ${C}_{1} , {C}_{2}$ are determined according to the initial conditions.

For instance if ${C}_{1} = \frac{1}{\sqrt{5}}$ and ${C}_{2} = - \frac{1}{\sqrt{5}}$ we will get the result shown in the question.