Question

Question #97138

Answer 1

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)-Clambda^(n+1)-Clambda^n=C lambda^n(lambda^2-lambda-1)=0` so the values

`lambda = 0, lambda = (1pmsqrt(5))/2` are eventual solutions for the difference equation.

So `a_n = C_0 0^n + C_1 ((1+sqrt(5))/2)^n+C_2 ((1-sqrt(5))/2)^n = C_1 ((1+sqrt(5))/2)^n+C_2 ((1-sqrt(5))/2)^n`

Now, `C_1,C_2` are determined according to the initial conditions.

For instance if `C_1 = 1/sqrt(5)` and `C_2 = -1/sqrt(5)` we will get the result shown in the question.