The partial fraction expansion of X(z) z which is

F(z) =	X(z)	=	z + 2
	z		z(2z2 – 7z + 3)

=	z + 2
	2z		z –	1		(z + 3)
				2

=

A0	+	A1			A2
z		z –	1		z – 3
			2

where,
A₀ = zF(z)|_z = 0

=	z + 2						|z = 0 =	2
	2		z –	1		(z – 3)		3
				2

A1 =		z –	1		F(z)|z = 1/2
			2

=	z + 2	|z = 1/2 = – 1
	2z(z – 3)

A₂ = (z – 3)F(z)|_{z = 3}

=	z + 2						|z = 3 =	1
	2z		z –	1				3
				2

Hence, by multiplying X(z)/z by z, we obtain

X(z) =	2	–	z			+	1	z
	3		z –	1			3	z – 3
				2

In the region |z| > 3 both poles are interior i.e., x(n) is causal.
Here, in the given function X(z) has two poles,

P1 =	1
	2

and P₂ = 3

∴ x(n) =	2	δ(n) –		1		n	u(n) +	1	(3)n u(n)
	3			2				3

Hence, alternative (D) is the correct choice.

Correct Option: D

The partial fraction expansion of X(z) z which is

F(z) =	X(z)	=	z + 2
	z		z(2z2 – 7z + 3)

=	z + 2
	2z		z –	1		(z + 3)
				2

=

A0	+	A1			A2
z		z –	1		z – 3
			2

where,
A₀ = zF(z)|_z = 0

=	z + 2						|z = 0 =	2
	2		z –	1		(z – 3)		3
				2

A1 =		z –	1		F(z)|z = 1/2
			2

=	z + 2	|z = 1/2 = – 1
	2z(z – 3)

A₂ = (z – 3)F(z)|_{z = 3}

=	z + 2						|z = 3 =	1
	2z		z –	1				3
				2

Hence, by multiplying X(z)/z by z, we obtain

X(z) =	2	–	z			+	1	z
	3		z –	1			3	z – 3
				2

In the region |z| > 3 both poles are interior i.e., x(n) is causal.
Here, in the given function X(z) has two poles,

P1 =	1
	2

and P₂ = 3

∴ x(n) =	2	δ(n) –		1		n	u(n) +	1	(3)n u(n)
	3			2				3

Hence, alternative (D) is the correct choice.