Exercise 4.4.23* (Most general sequence such that $u_n = 5u_{n-1} - 6 u_{n-2} +n$)

Proof.

(a)

The associate characteristic polynomial is $$p(x)=x^2-5x+6=(x-2)(x-3).$$

By Theorem 4.10, there are complex constants $\alpha ,\beta$ such that

$$u_n=\alpha 2^n+\beta 3^n.$$

Conversely, since $2^2=5\cdot 2-6$ and $3^2=5\cdot 3-6$, we obtain for all $n\ge 2$,

$$\begin{array}{lll}\hfill 2^n=5\cdot 2^{n-1}-6\cdot 2^{n-2},& & \hfill \\ \hfill 3^n=5\cdot 3^{n-1}-6\cdot 3^{n-2}.& & \hfill \end{array}$$

Therefore the sequence ${(u_n)}_{n\in \mathbb{N}}$ defined by $u_n=\alpha 2^n+\beta 3^n$ satisfies for all $n\in \mathbb{N}$,

$$u_n=5u_{n-1}-6u_{n-2}.$$

The most general sequence of complex numbers ${(u_n)}_{n\in \mathbb{N}}$ such that, for $n\ge 2$, $u_n=5u_{n-1}-6u_{n-2}$ is defined by

$$u_n=\alpha 2^n+\beta 3^n(n\in \mathbb{N})$$

for some constant complex values $\alpha ,\beta$.

Then

is true for $\gamma =1∕2$, because the solution of $-\gamma =-5\gamma +6\gamma =1$ is $\gamma =1∕2$. By part (a), $u_n-\gamma =\alpha 2^n+\beta 3^n$ for some constant $\alpha ,\beta$, so

Conversely, the sequence defined by (2) satisfies $u_n=5u_{n-1}-6u_{n-2}+1$ for all $n\ge 2$. (c) Suppose that for all $n\ge 2$,

Then $\lambda ,\mu$ satisfy

if for all $n\ge 2$

or equivalently,

that is, if $(\lambda ,\mu )$ is solution of the system

which gives $\lambda =-1∕2,\mu =-7∕4$.

By part (a), there are constant $\alpha ,\beta$ such that

Conversely, if (3) is true, then for all $n\ge 2$,

Then, using part (a),

The most general sequence of complex numbers ${(u_n)}_{n\in \mathbb{N}}$ such that, for $n\ge 2$, $u_n=5u_{n-1}-6u_{n-2}+n$ is defined by

for some constant complex values $\alpha ,\beta$. □