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marshall27 [118]
3 years ago
9

An employee joined a company in 2009 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 2009.
b) What will the salary of this employee be in 2017?
c) Find an explicit formula for the salary of this employee n years after 2009.
Mathematics
1 answer:
stepladder [879]3 years ago
6 0

Answer:

(a) The required recurrence relation for  the salary of the employee of n years after 2009 is a_n=1.05a_{n-1}+1000.

(b)The salary of the employee will be $83421.88 in 2017.

(c) \therefore a_n=70,000 . \ 1.05^n-20,000

Step-by-step explanation:

Summation of a G.P series

\sum_{i=0}^n r^i= \frac{r^{n+1}-1}{r-1}

(a)

Every year the salary is increasing 5% of the salary of the previous year plus $1000.

Let a_n represents the salary of the employee of n years after 2009.

Then a_{n-1} represents the salary of the employee of (n-1) years after 2009.

Then a_n= a_{n-1}+5\%.a_{n-1}+1000

             =a_{n-1}+0.05a_{n-1}+1000

             =(1+0.05)a_{n-1}+1000

            =1.05a_{n-1}+1000

The required recurrence relation for  the salary of the employee of n years after 2009 is a_n=1.05a_{n-1}+1000.

(b)

Given, a_0=\$50,000

a_n=1.05a_{n-1}+1000

Since 2017 is 8 years after 2009.

So, n=8.

∴ a_8

=1.05 a_7+1000

=1.05(1.05a_6+1000)+1000

=1.05^2a_6+1.05\times 1000+1000

=1.05^2(1.05a_5+1000)+1.05\times 1000+1000

=1.05^3a_5+1.05^2\times 1000+1.05\times 1000+1000

=1.05^3(1.05a_4+1000)+1.05^2\times 1000+1.05\times 1000+1000

=1.05^4a_4+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^4(1.05a_3+1000)+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^5a_3+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^5(1.05a_2+1000)+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^6a_2+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^6(1.05a_1+1000)+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^7a_1+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^7(1.05a_0+1000)+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^8a_0+1.05^7\times1000+1.05^6\times1000+1.05^51000+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000

=1.05^8a_0+(1.05^7+1.05^6+1.05^5+1.05^4+1.05^3+1.05^2+1.05+1)1000

=1.05^8 \times 50,000+\frac{1.05^8-1}{1.05-1}\times 1000

=1.05^8\times 50,000+20,000(1.58^8-1)

=70,000\times 1.05^8-20,000

≈$83421.88

The salary of the employee will be $83421.88 in 2017.

(c)

Given, a_0=\$50,000

a_n=1.05a_{n-1}+1000

We successively apply the recurrence relation

a_n=1.05a_{n-1}+1000

    =1.05^1a_{n-1}+1.05^0.1000

   =1.05^1(1.05a_{n-2}+1000)+1.05^0.1000

   =1.05^2a_{n-2}+1.05^1.1000+1.05^0.1000

   =1.05^2(1.05a_{n-3}+1000)+(1.05^1.1000+1.05^0.1000)

   =1.05^3a_{n-3}+(1.05^2.1000+1.05^1.1000+1.05^0.1000)

                    ...............................

                   .................................

  =1.05^na_{n-n}+\sum_{i=0}^{n-1}1.05^i.1000

 =1.05^na_0+1000\sum_{i=0}^{n-1}1.05^i

 =1.05^n.50,000+1000.\frac{1.05^n-1}{1.05-1}

 =1.05^n.50,000+20,000.(1.05^n-1)

 =(50,000+20,000)1.05^n-20,000

 =70,000 . \ 1.05^n-20,000

\therefore a_n=70,000 . \ 1.05^n-20,000

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The attached figure

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