Answer:
The amount of money in Victoria's account at the end of n-th year will be,
$ [tex] (965 \times (1.02)^{(n -1)} - 765)[/tex]
Step-by-step explanation:
The amount of money in Victoria's account at the end of n-th year will be,
$ [tex](200\times (1.02)^{(n-1)} + 15 \times ((1.02)^{(n-1)} + (1.02)^{(n -2)} + (1.02)^{(n-3)} + ........ + (1.02)))[/tex]
= $ [tex](200 \times (1.02)^{(n -1)} + 15 \times \frac{1.02 \times ((1.02)^{(n-1)} - 1)}{0.02})[/tex]
= $ [tex](200 \times (1.02)^{(n -1)} + 15 \times 51 \times ((1.02)^{(n-1)} - 1))[/tex]
= $ [tex] (965 \times (1.02)^{(n -1)} - 765)[/tex]
Since, the interest is compounded over that $ 200 (which was there in the account at the end of 1st year) for (n-1) years and for (n-1) years, (n-2) years, (n -3) years, ...... 1 year respectively over those $ 15 which are deposited in the account at the end of 1 , 2, 3, ...... (n-1) years.
Answer:
[tex]a_{n}=\left [ a\left ( n-1 \right )+15 \right ]*1.02\\[/tex]
Step-by-step explanation:
1) A Recursive formula always makes reference to the previous term, written as a function.
2) From $200 to $215 there was a 7.5% growth, so since the question states that from the beginning of the second year there will be a regular growth.
3) Since it is 2% of interest compounded annually q=1+ 0.02.
[tex]A=215(1+\frac{0.02}{1})^{1}\\A=219.3[/tex]
Or simply
215*1.02=219.3
4) We can write a Recursive formula this way:
[tex]a(n)=215[/tex]
[tex]a(n)=a(n-1)(1.02)[/tex]
5) But since we have a pattern, She will deposit $15 yearly then we must make a little adjustment then add $15. to that. Considering the first term to be 200
[tex]a_{n}=\left [ a\left ( n-1 \right )+15 \right ]*1.02\\a_{1}=200[/tex]
