**Given:**

The population of catfish after *n* months is
Pn=1.08Pn−1−300
. (1)

Initial population,
P0=5000
.

**Calculation:**

Obtain the value of
P1
.

Substitute 1 for *n* in equation (1),

P1=1.08P1−1−300

P1=1.08P0−300
(2)

Substitute 5000 for
P0
in equation (2),

P1=1.08(5000)−300=5400−300=5100

Thus, the value of
P1=5100
.

Obtain the value of
P2
.

Substitute 2 for *n* in equation (1),

P2=1.08P2−1−300

P2=1.08P1−300
(3)

Substitute 5100 for
P1
in equation (3),

P2=1.08(5100)−300=5508−300=5208

Thus, the value of
P2=5208
.

Obtain the value of
P3
.

Substitute 3 for *n* in equation (1),

P3=1.08P3−1−300

P3=1.08P2−300
(4)

Substitute 5208 for
P2
in equation (4),

P3=1.08(5208)−300=5624.64−300=5324.64

Thus, the value of
P3=5325
.

Obtain the value of
P4
.

Substitute 4 for *n* in equation (1),

P4=1.08P4−1−300

P4=1.08P3−300
(5)

Substitute 5325 for
P3
in equation (4),

P4=1.08(5325)−300=5751−300=5451

Thus, the value of
P4=5451
.

Obtain the value of
P5
.

Substitute 5 for *n* in equation (1),

P5=1.08P5−1−300

P5=1.08P4−300
(6)

Substitute 5451 for
P4
in equation (6),

P5=1.08(5451)−300=5887.08−300=5587.08

Thus, the value of
P5=5587
.

Obtain the value of
P6
.

Substitute 6 for *n* in equation (1),

P6=1.08P6−1−300

P6=1.08P5−300
(7)

Substitute 5587 for
P5
in equation (6),

P6=1.08(5587)−300=6033.96−300=5733.96

Thus, the value of
P6=5734
.

Therefore, the number of catfish in the pond after six months is 5734.