Difference equation x(k+2)+3x(k+1)+2x(k)=0

Take the z-transform of both sides using the shifting theorems:

$\mathcal{Z}[x(k+1)] = zX(z) - zx(0)$, $\quad\mathcal{Z}[x(k+2)] = z^2X(z) - z^2x(0) - zx(1)$

Step 1: Transform

$$[z^2X(z) - z^2x(0) - zx(1)] + 3[zX(z) - zx(0)] + 2X(z) = 0$$

Substituting $x(0)=0$, $x(1)=1$:

$$\begin{aligned} z^2X(z) - z + 3zX(z) + 2X(z) &= 0 \\ (z^2+3z+2)X(z) &= z \\ X(z) &= \frac{z}{(z+1)(z+2)} \end{aligned}$$

Step 2: Partial fractions of $X(z)/z$

$$\frac{X(z)}{z} = \frac{1}{(z+1)(z+2)} = \frac{A}{z+1} + \frac{B}{z+2}$$

$A = \left.\frac{1}{z+2}\right|_{z=-1} = 1$, $\quad B = \left.\frac{1}{z+1}\right|_{z=-2} = -1$

$$X(z) = \frac{z}{z+1} - \frac{z}{z+2}$$

Step 3: Inverse z-transform

Using $\mathcal{Z}^{-1}\left[\frac{z}{z-a}\right] = a^k$:

$$x(k) = (-1)^k - (-2)^k, \quad k = 0, 1, 2, \dots$$

Check

Both agree, and the initial conditions are satisfied. The pole at $z=-2$ lies outside the unit circle, so the sequence grows in magnitude with alternating sign.

Answer: $x(k) = (-1)^k - (-2)^k$, i.e. $x(k) = 0, 1, -3, 7, -15, 31, \dots$