Answer:
[tex]f(n) = \frac{4^n-0.25}{3}[/tex] --- Job offer 1
[tex]g(n) = 16n[/tex] --- Job offer 2
[tex]1 \le n \le 10[/tex]
Step-by-step explanation:
Given
Job offer 1:
Offer = $0.25
[tex]Day 1 = \$1\ Day 2 = \$4\ Day 3 = \$16......[/tex]
Job offer 2:
Daily = $16
Required
Determine the equation for both jobs
For job offer 1:
Considering the pay for day 1, day 2,....
The sequence shows a geometry progression where the payment between subsequent days is a product of 4 by the payment of the previous day.
The sequence can be represented as:
[tex]T_1 = 1; T_2 = 4; T_3 = 16[/tex]
The common ratio (r) is:
[tex]r = \frac{T_2}{T_1} = \frac{4}{1} = 4[/tex]
The sum of n terms is the total salary received in n days.
[tex]S_n = \frac{a(r^n - 1)}{r-1}[/tex]
[tex]S_n = \frac{1 * (4^n - 1)}{4-1}[/tex]
[tex]S_n = \frac{4^n - 1}{3}[/tex]
So, the equation for job offer 1 is:
[tex]f(n) = Offer + S_n[/tex]
[tex]f(n) = 0.25+ \frac{4^n - 1}{3}[/tex]
Take LCM
[tex]f(n) = \frac{0.75 + 4^n - 1}{3}[/tex]
Collect like terms
[tex]f(n) = \frac{4^n - 1+0.75}{3}[/tex]
[tex]f(n) = \frac{4^n-0.25}{3}[/tex]
For job offer 2:
Daily payment of $16 implies that job offer 2 pays 16n for n days.
So:
[tex]g(n) = 16n[/tex]
In both cases: [tex]1 \le n \le 10[/tex]
