Showing that $x_n=nx^n$ is a solution of the recurrence relation

Question

Given that $r^2 - ar - b$ has the double root $x$, how do you show that $x_n = nx^n$ is a solution of $x_{n+1}=-4x_n-4x_{n-1}$? I can show $x_n$ = $x^n$ but I just can't show for $x_n$ = $nx^n$.

Answer 1

Let the characteristic equation $f(x)$ be $$f(x)=x^n-c_1x^{n-1}-\cdots -c_kx^{n-k}$$

Since $\alpha$ is a double root, we have $$f(\alpha)=f'(\alpha)=0$$ This gives $$\alpha f'(\alpha) = n\alpha^n-c_1(n-1)\alpha^{n-1} - \cdots -c_k(n-k)\alpha^{n-k}=0$$

Therefore, $n \alpha^n$ is a solution. A triple root, etc can be proved like this as well.