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

(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`

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`