x(k) for k = 0 to 4 from X(z)

The first few values are found by direct (long) division. A closed form from partial fractions confirms them.

Method 1: Long division

$$X(z) = \frac{10z + 5}{(z-1)(z-2)} = \frac{10z + 5}{z^2 - 3z + 2} = \frac{10z^{-1} + 5z^{-2}}{1 - 3z^{-1} + 2z^{-2}}$$

Divide $10z^{-1} + 5z^{-2}$ by $1 - 3z^{-1} + 2z^{-2}$:

$$X(z) = 0 + 10z^{-1} + 35z^{-2} + 85z^{-3} + 185z^{-4} + \dots$$

Equivalently, the recursion $x(k) = 3x(k-1) - 2x(k-2)$ holds for $k \ge 3$: for example $x(3) = 3(35) - 2(10) = 85$ and $x(4) = 3(85) - 2(35) = 185$.

Method 2: Closed form (check)

$$\frac{X(z)}{z} = \frac{10z+5}{z(z-1)(z-2)} = \frac{A}{z} + \frac{B}{z-1} + \frac{C}{z-2}$$ $$A = \frac{5}{(-1)(-2)} = 2.5, \quad B = \frac{15}{(1)(-1)} = -15, \quad C = \frac{25}{(2)(1)} = 12.5$$ $$X(z) = 2.5 - \frac{15z}{z-1} + \frac{12.5z}{z-2}$$ $$x(k) = 2.5\,\delta(k) - 15 + 12.5(2)^k$$

• $x(0) = 2.5 - 15 + 12.5 = 0$
• $x(1) = -15 + 25 = 10$
• $x(2) = -15 + 50 = 35$
• $x(3) = -15 + 100 = 85$
• $x(4) = -15 + 200 = 185$

Result

The pole at $z = 2$ lies outside the unit circle, so $x(k)$ grows without bound.

Answer: $x(0) = 0$, $x(1) = 10$, $x(2) = 35$, $x(3) = 85$, $x(4) = 185$. In general $x(k) = -15 + 12.5(2)^k$ for $k \ge 1$.