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 em- ployee 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 em- ployee n years after 2017.

  a) s[n] = 1.05s[n-1] +1000; s[0] = 50,000

  b) $83,421.88

  c) s[n] = 70000·1.05^n -20000

Step-by-step explanation:

a) Let s[n] represents the employee's salary in dollars n years after 2017. The salary in 2017 is given as $50,000, so ...

  s[0] = 50000 . . . . the recursion relation initial condition

The next year, the salary is multiplied by 1.05 and 1000 is added:

  s[1] = 1.05·s[0] +1000

This pattern repeats, so the recursion relation is ...

  s[n] = 1.05·s[n-1] +1000

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c) It is convenient to find a formula for the salary before trying to compute the salary 8 years on. So, we work part (c) before part (b).

The base salary gets multiplied by 1.05 each year, so can be described by the exponential function ...

  base salary after n years = 50,000·1.05^n

The add-on to the raise becomes a geometric sequence whose common ratio is 1.05. The sum can be described by the formula for the sum of such a sequence:

  sum = a1(r^n -1)/(r -1) . . . . . (see note below)

  sum of add-ons = 1000(1.05^n -1)/(1.05 -1) = 20000(1.05^n -1)

So, the total salary after n years is ...

  s[n] = 50000·1.05^n + 20000(1.05^n -1)

The exponential terms can be combined, so we have the explicit formula ...

  s[n] = 70000·1.05^n -20000

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b) The year 2025 is 8 years after 2017, so we want to find s[8].

  s[8] = 70000·1.05^8 -20000 = 70000·1.4774554 -20000

     = 103,421.88 -20000 = 83,421.88

In 2025, the employee's salary will be $83,421.88.

Note on the sum of the add-ons

A geometric sequence starts with term a1 and gets multiplied by a ratio r to make the next term: a2=r·a1; a3=r²·a1,

The sum of such a geometric sequence has the formula

  Sn = a1·(r^n -1)/(r -1)

Here, the sequence starts in year 1 with a1=1000 and has a common ratio r=1.05. The denominator of the sum is r-1 = 0.05 = 1/20, so the sum can be written as

  Sn = 20000·(1.05^n -1).

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step-by-step explanation:

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