Respuesta :
Answer:
(a) The required recurrence relation for the salary of the employee of n years after 2009 is [tex]a_n=1.05a_{n-1}+1000[/tex].
(b)The salary of the employee will be $83421.88 in 2017.
(c) [tex]\therefore a_n=70,000 . \ 1.05^n-20,000[/tex]
Step-by-step explanation:
Summation of a G.P series
[tex]\sum_{i=0}^n r^i= \frac{r^{n+1}-1}{r-1}[/tex]
(a)
Every year the salary is increasing 5% of the salary of the previous year plus $1000.
Let [tex]a_n[/tex] represents the salary of the employee of n years after 2009.
Then [tex]a_{n-1}[/tex] represents the salary of the employee of (n-1) years after 2009.
Then [tex]a_n= a_{n-1}+5\%.a_{n-1}+1000[/tex]
[tex]=a_{n-1}+0.05a_{n-1}+1000[/tex]
[tex]=(1+0.05)a_{n-1}+1000[/tex]
[tex]=1.05a_{n-1}+1000[/tex]
The required recurrence relation for the salary of the employee of n years after 2009 is [tex]a_n=1.05a_{n-1}+1000[/tex].
(b)
Given, [tex]a_0=\$50,000[/tex]
[tex]a_n=1.05a_{n-1}+1000[/tex]
Since 2017 is 8 years after 2009.
So, n=8.
∴ [tex]a_8[/tex]
[tex]=1.05 a_7+1000[/tex]
[tex]=1.05(1.05a_6+1000)+1000[/tex]
[tex]=1.05^2a_6+1.05\times 1000+1000[/tex]
[tex]=1.05^2(1.05a_5+1000)+1.05\times 1000+1000[/tex]
[tex]=1.05^3a_5+1.05^2\times 1000+1.05\times 1000+1000[/tex]
[tex]=1.05^3(1.05a_4+1000)+1.05^2\times 1000+1.05\times 1000+1000[/tex]
[tex]=1.05^4a_4+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000[/tex]
[tex]=1.05^4(1.05a_3+1000)+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000[/tex]
[tex]=1.05^5a_3+1.05^4\times1000+1.05^3\times 1000+1.05^2\times 1000+1.05\times 1000+1000[/tex]
[tex]=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[/tex]
[tex]=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[/tex]
[tex]=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[/tex]
[tex]=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[/tex]
[tex]=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[/tex]
[tex]=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[/tex]
[tex]=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[/tex]
[tex]=1.05^8 \times 50,000+\frac{1.05^8-1}{1.05-1}\times 1000[/tex]
[tex]=1.05^8\times 50,000+20,000(1.58^8-1)[/tex]
[tex]=70,000\times 1.05^8-20,000[/tex]
≈$83421.88
The salary of the employee will be $83421.88 in 2017.
(c)
Given, [tex]a_0=\$50,000[/tex]
[tex]a_n=1.05a_{n-1}+1000[/tex]
We successively apply the recurrence relation
[tex]a_n=1.05a_{n-1}+1000[/tex]
[tex]=1.05^1a_{n-1}+1.05^0.1000[/tex]
[tex]=1.05^1(1.05a_{n-2}+1000)+1.05^0.1000[/tex]
[tex]=1.05^2a_{n-2}+1.05^1.1000+1.05^0.1000[/tex]
[tex]=1.05^2(1.05a_{n-3}+1000)+(1.05^1.1000+1.05^0.1000)[/tex]
[tex]=1.05^3a_{n-3}+(1.05^2.1000+1.05^1.1000+1.05^0.1000)[/tex]
...............................
.................................
[tex]=1.05^na_{n-n}+\sum_{i=0}^{n-1}1.05^i.1000[/tex]
[tex]=1.05^na_0+1000\sum_{i=0}^{n-1}1.05^i[/tex]
[tex]=1.05^n.50,000+1000.\frac{1.05^n-1}{1.05-1}[/tex]
[tex]=1.05^n.50,000+20,000.(1.05^n-1)[/tex]
[tex]=(50,000+20,000)1.05^n-20,000[/tex]
[tex]=70,000 . \ 1.05^n-20,000[/tex]
[tex]\therefore a_n=70,000 . \ 1.05^n-20,000[/tex]
