Question

22. An employee joined a company in 2017 with a starting salary of $50,000. Every year this employee receives a raise of $1000 plus 5% of the salary of the previous year. a) Set up a recurrence relation for the salary of this employee n years after 2017. b) What will the salary of this employee be in 2025? c) Find an explicit formula for the salary of this employee n years after 2017.

Answer 1

Answer:

(a) The recurrence relation for the salary is

$S_{n+1}=1.05*S_n+1000\\\\S_0=50000$

(b) The salary 25 years after 2017 will be $217044.85.

(c) $S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n$

Step-by-step explanation:

We can define the next year salary $S_{n+1}$ as

$S_{n+1}=S_n+1000+0.05*S_n=1.05*S_n+1000$

wit S0=$50000

If we extend this to 2 years from 2017 (n+2), we have

$S_{n+2}=1.05*S_{n+1}+1000=1.05*(1.05*S_n+1000)+1000\\S_{n+2} =1.05^2*S_n+1.05*1000+1000\\S_{n+2}=1.05^2*S_n+1000*(1.05^1+1)$

Extending to 3 years (n+3)

$S_{n+3}=1.05*S_{n+2}+1000=1.05(1.05^2*S_n+1000*(1.05^1+1))+1000\\\\S_{n+3}=1.05^3S_n+1.05*1000*(1.05^1+1)+1000\\\\S_{n+3}=1.05^3*S_n+1000*(1.05^2+1.05^1+1)$

Extending to 4 years (n+4)

$S_{n+4}=1.05*S_{n+3}+1000=1.05*(1.05^3*S_n+1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1.05*1000*(1.05^2+1.05^1+1))+1000\\\\S_{n+4}=1.05^4S_n+1000*(1.05^3+1.05^2+1.05^1+1.05^0)$

We can now express a general equation for S_n (salary at n years from 2017)

$S_n=1.05^nS_0+1000*\sum_{0}^{n-1}1.05^n$

The salary at 25 years from 2017 (n=25) will be

$S_{25}=1.05^{25}S_0+1000*\sum_{0}^{24}1.05^i\\\\S_{25}=3.386*50000+1000*47.72=217044.85$