a high - interest savings account pays 5.5% interest compounded annually. if $300 is deposited initially and…

Answer:

The formula for the future - value of an ordinary annuity compounded annually is $F = A\frac{(1 + r)^{n}-1}{r}$, where $A$ is the annual payment, $r$ is the interest rate per period, and $n$ is the number of periods. Here, $A = 300$, $r=0.055$, and $n = 10$.

We can also think of it in terms of a sum of compound - interest values.

The first deposit of $300$ will earn interest for 10 years, the second deposit for 9 years, and so on.

The value of the first deposit after 10 years is $300(1 + 0.055)^{10}$, the value of the second deposit after 9 years is $300(1 + 0.055)^{9}$, and in general, the value of the $k^{th}$ deposit (where $k$ ranges from 1 to 10) after $(10-(k - 1))$ years is $300(1+0.055)^{10-(k - 1)}$.

The sum of the money in the account after 10 years is a geometric series.

The general term of a geometric series is $a\cdot r^{n - 1}$. Here, $a$ is the first - term and $r$ is the common ratio.

The first deposit of $300$ made at the start of the first year will have a value of $300(1 + 0.055)^{10}$ at the end of 10 years. The second $300$ deposit made at the start of the second year will have a value of $300(1 + 0.055)^{9}$ at the end of 10 years, and so on.

The sum of the money in the account after 10 years is $\sum_{n = 1}^{10}300(1 + 0.055)^{n-1}\times(1 + 0.055)^{10-(n - 1)}=\sum_{n = 1}^{10}300(1.055)^{n - 1}\times(1.055)^{10-(n - 1)}=\sum_{n = 1}^{10}300(1.055)^{9+(n - 1-(n - 1))}=\sum_{n = 1}^{10}300(1.055)^{n - 1}\times(1.055)^{9}=\sum_{n = 1}^{10}300\times1.055\times(1.055)^{n - 1}=\sum_{n = 1}^{10}315(1.055)^{n - 1}$

Let's analyze the problem step - by - step:

Step1: Recall the compound - interest formula

The compound - interest formula for a single deposit $P$ after $t$ years with an annual interest rate $r$ compounded annually is $A=P(1 + r)^{t}$.

Step2: Analyze each deposit

The first $300$ deposit earns interest for 10 years, so its value is $300(1 + 0.055)^{10}$. The second $300$ deposit (made at the start of the second year) earns interest for 9 years, so its value is $300(1 + 0.055)^{9}$, and in general, the $n^{th}$ deposit (where $n$ ranges from 1 to 10) earns interest for $(10-(n - 1))$ years, and its value is $300(1 + 0.055)^{10-(n - 1)}$.

Step3: Write the sum as a geometric series

The sum of the values of all the deposits is a geometric series $\sum_{n = 1}^{10}300(1 + 0.055)^{n - 1}\times(1 + 0.055)^{10-(n - 1)}$. Simplifying this sum gives us a geometric series of the form $\sum_{n = 1}^{10}300\times1.055\times(1.055)^{n - 1}=\sum_{n = 1}^{10}315(1.055)^{n - 1}$

The correct sum is $\sum_{n = 1}^{10}300(1.055)^{n - 1}\times1.055=\sum_{n = 1}^{10}315(1.055)^{n - 1}$

However, if we consider the first - term of the geometric series in the standard form $a\cdot r^{n - 1}$, when we rewrite the sum of the money in the account, we know that the first deposit of $300$ after 10 years of compounding at $5.5%$ and considering the series structure:

The sum of the money in the account after 10 years is $\sum_{n = 1}^{10}300(1.055)\times(1.055)^{n - 1}=\sum_{n = 1}^{10}315(1.055)^{n - 1}$

If we assume there is a small error in the problem setup and we consider the following:

The first deposit of $300$ made at the start of the first year. The amount in the account after 10 years for this deposit is $300(1 + 0.055)^{10}$. The second deposit of $300$ made at the start of the second year has a value of $300(1 + 0.055)^{9}$ at the end of 10 years.

The sum of the amounts is $\sum_{n = 1}^{10}300(1.055)^{n - 1}\times(1.055)^{10-(n - 1)}$.

We know that for a geometric series $S=\sum_{n = 1}^{N}a\cdot r^{n - 1}$, where $a$ is the first - term and $r$ is the common ratio.

The first deposit of $300$ made at $t = 0$ will have a value of $300(1.055)^{10}$ at $t = 10$, the second deposit of $300$ made at $t = 1$ will have a value of $300(1.055)^{9}$ at $t = 10$.

The sum of the money in the account is $\sum_{n = 1}^{10}300\times1.055\times(1.055)^{n - 1}=\sum_{n = 1}^{10}315(1.055)^{n - 1}$

The closest correct form among the given options (assuming some possible mis - typing in the problem or options) is: There is an error in the given options. But if we assume the first - term of the geometric series is calculated as follows: The first deposit of $300$ made at the start of the first year. After the first - year, the amount due to the first deposit is $300\times(1 + 0.055)=300\times1.055 = 315$. The sum of the money in the account after 10 years (using the geometric - series formula for the future value of an annuity - like situation) is $\sum_{n = 1}^{10}315(1.055)^{n - 1}$

If we assume that the $300$ deposit grows to $300\times(1.055)=315$ in the first year and then we consider the geometric series for the subsequent years of compounding, the sum of the money in the account after 10 years is $\sum_{n = 1}^{10}315(1.055)^{n - 1}$

If we assume a small error in the options and we consider the fact that the initial $300$ deposit earns $300\times0.055 = 16.5$ in the first year, so the amount becomes $300+16.5=316.5$ and the sum of the money in the account after 10 years is $\sum_{n = 1}^{10}316.5(1.055)^{n - 1}$

Answer:

$\sum_{n = 1}^{10}316.5(1.055)^{n - 1}$

Explanation:

Step1: Calculate the amount of the first - year deposit after one year

The initial $300$ deposit earns $300\times0.055=16.5$ in the first year. So the amount becomes $300 + 16.5=316.5$.

Step2: Recognize the geometric - series structure

The money in the account forms a geometric series. Each subsequent deposit's growth is based on the compound - interest formula. The general term of a geometric series is $a\cdot r^{n - 1}$, where $a = 316.5$ (the amount of the first - year deposit after one year) and $r=1.055$ (the growth factor due to the $5.5%$ interest rate). The sum of a geometric series for $n$ terms is $\sum_{n = 1}^{N}a\cdot r^{n - 1}$. Here $N = 10$.