Answer:
[tex]n(t) =5700+ 1000t[/tex]; [tex]t > 0[/tex]
[tex]n(t) = 1700[/tex]; [tex]t =0[/tex]
Step-by-step explanation:
Given
[tex]n(0) = 1700[/tex] --- initial
[tex]n(1) = 5000[/tex]
[tex]n(2) = 6000[/tex]
[tex]n(3) = 7000[/tex]
Required
Explicit formula
First, we calculate the difference between the earnings of successive months.
[tex]d = n(2) - n(1) = n(3) - n(2)[/tex]
So, we have:
[tex]d = 6000 - 5000[/tex]
[tex]d = 1000[/tex]
The explicit formula is:
[tex]n(t) = n(0) + n(1) + (t - 1)d[/tex]
Where
[tex]t \to[/tex] months
[tex]n(t) = 1700 + 5000 + (t - 1) * 1000[/tex]
Open bracket
[tex]n(t) = 1700 + 5000 + 1000t - 1000[/tex]
Collect like terms
[tex]n(t) = 1700 + 5000 - 1000+ 1000t[/tex]
[tex]n(t) =5700+ 1000t[/tex]
