[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{15\% of 50}}{\left( \cfrac{15}{100} \right)50}\implies (0.15)50[/tex]
year 1................... (0.15)50
year 2................. (0.15)(0.15)50
year 3................. (0.15)(0.15)(0.15)50
year n................. 50(0.15)ⁿ
when the next term is simply obtained by multiplying the current one by some multiplier, is a geometric sequence.
Answer:
50(1.15)^n-1
Step-by-step explanation:
1. Let's consider the first three terms of g(n) to get a sense of how the function values change as n increases.
2. The first term is Luiza's account balance at the first year of the saving, which is the initial amount she deposited. We know this to be $50.
The second term is Luiza's account balance at the second year. Since the account accumulated 15% each year, it was 1.15 times the balance in the first year, which is $50*1.15=$57.50.
The third term is Luiza's account balance at the third year. Again, this is 1.15 times the balance of the year before that, which is $57.50*1.15=$66.125.
To summarize:
g(1)=50 g(2)=50*1.15 g(3)=50*1.15*1.15
We can see that each term is 1.15 times its preceding term. There is a constant ratio between consecutive terms. Therefore, this is a geometric sequence.
3. We can write an explicit formula for this geometric sequence using the form A*B^n-1. In this form, A, is the first term and B is the common ratio. What are the appropriate values for our case?
4. In conclusion, g is a geometric sequence.
An explicit formula for the sequence is g(n)=50*1.15^n-1
Note that this solution strategy results in this formula; however, an equally correct solution can be written in other equivalent forms as well.
