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.