Respuesta :
Answer:
(a) The recurrence relation for the salary is
[tex]S_{n+1}=1.05*S_n+1000\\\\S_0=50000[/tex]
(b) The salary 25 years after 2017 will be $217044.85.
(c) [tex]S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n[/tex]
Step-by-step explanation:
We can define the next year salary [tex]S_{n+1}[/tex] as
[tex]S_{n+1}=S_n+1000+0.05*S_n=1.05*S_n+1000[/tex]
wit S0=$50000
If we extend this to 2 years from 2017 (n+2), we have
[tex]S_{n+2}=1.05*S_{n+1}+1000=1.05*(1.05*S_n+1000)+1000\\S_{n+2} =1.05^2*S_n+1.05*1000+1000\\S_{n+2}=1.05^2*S_n+1000*(1.05^1+1)[/tex]
Extending to 3 years (n+3)
[tex]S_{n+3}=1.05*S_{n+2}+1000=1.05(1.05^2*S_n+1000*(1.05^1+1))+1000\\\\S_{n+3}=1.05^3S_n+1.05*1000*(1.05^1+1)+1000\\\\S_{n+3}=1.05^3*S_n+1000*(1.05^2+1.05^1+1)[/tex]
Extending to 4 years (n+4)
[tex]S_{n+4}=1.05*S_{n+3}+1000=1.05*(1.05^3*S_n+1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1.05*1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1000*(1.05^3+1.05^2+1.05^1+1.05^0)[/tex]
We can now express a general equation for S_n (salary at n years from 2017)
[tex]S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n[/tex]
The salary at 25 years from 2017 (n=25) will be
[tex]S_{25}=1.05^{25}S_0+1000*\sum_{0}^{24}1.05^i\\\\S_{25}=3.386*50000+1000*47.72=217044.85[/tex]
