Answer:
Step-by-step explanation:
(a)
From the given information:
Let An denote the amount in the account at the end of n month
Thus,
[tex]A_o[/tex] =1000
After the first month ends, interest = [tex]=A_o \times \dfrac{3}{100} \times \dfrac{1}{12}[/tex]
= [tex]0.0025A_o[/tex]
Additional deposit = $200
Now,
[tex]A_1 = A_o + 0.0025A_o + 200[/tex]
[tex]=1.0025A_o + 200 --- eqn(1)[/tex]
Interest after 2nd month, = [tex]A_1 \times \dfrac{3}{100} \times \dfrac{1}{2}[/tex]
Additional deposit = $200
[tex]A_2 = A_1 + 0.0025A_1 + 200[/tex]
[tex]=1.0025A_1 + 200 --- eqn(2)[/tex]
Thus:
[tex]A_n = 1.0025A_{n-1} +200[/tex]
(b)
From above;
[tex]A_1 = 1.0025A_ 0+ 200[/tex]
Thus;
[tex]A _2 = 1.0025A_ 1+ 200[/tex]
[tex]A_2 = 1.0025( 1.0025A_ 0+ 200)+ 200[/tex]
[tex]A_2 = 1.0025^2A_o +200 ( 1+1.0025)[/tex]
[tex]A _3 = 1.0025A_ 2+ 200[/tex]
[tex]A _3 = 1.0025( 1.0025^2A_o +200 ( 1+1.0025))+ 200[/tex]
[tex]A _3 = 11.0025^3A_o +200 ( 1+1.0025 + 1.0025^2)[/tex]
To continuity...
[tex]A_n = 1.0025^n A_o + 200 (1+1.0025 +1.0025^2 +...+ 1.0025^{n-1} ) ---- (3)[/tex]
Thus, we have a geometric term [tex]1 + 1.0025 +1.0025^2 +...+ 1.0025^{n-1}[/tex]
where:
a = 1 (first term)
r = 1.0025>1 (common ratio)
Thus;
[tex]1 + 1.0025 +1.0025^2 +...+ 1.0025^{n-1} = \dfrac{a(r^n -1)}{r-1}[/tex]
[tex]= \dfrac{1(1.0025^n -1)}{1.0025-1}[/tex]
[tex]= 400(1.0025^n -1)[/tex]
From equation(3)
[tex]A_n = (1.0025^n \times 1000) + 80000(1.0025^n -1)[/tex]
[tex]A_n = 1.0025^n \times 81000-80000[/tex]
