a high - interest savings account pays 5.5% i...

# Explanation: ## Step1: Recall compound - interest formula for each deposit The compound - interest formula for a single deposit $P$ with an annual interest rate $r$ compounded annually after $t$ years is $A = P(1 + r)^t$. The first deposit of $P_1=300$ will earn interest for 10 years, the second deposit of $P_2 = 300$ will earn interest for 9 years, and in general, the $n$th deposit (where the first deposit is $n = 1$) will earn interest for $10-(n - 1)$ years. Here, $r=0.055$. ## Step2: Write out the sum of the future - values of each deposit The future - value of the $n$th deposit of $300$ made at the start of the $n$th year is $A_n=300(1 + 0.055)^{10-(n - 1)}=300(1.055)^{11 - n}$. We can rewrite the index of summation. Let $k=11 - n$. When $n = 1$, $k = 10$ and as $n$ goes from 1 to 10, $k$ goes from 10 to 1. Another way is to consider the general form of a geometric series. The first deposit of $300$ after 10 years has value $300(1.055)^{10}$, the second deposit of $300$ after 9 years has value $300(1.055)^{9}$, etc. The sum of the money in the account after 10 years is a geometric series $\sum_{n = 1}^{10}300(1.055)^{10-(n - 1)}=\sum_{n = 1}^{10}300(1.055)^{11 - n}$. If we rewrite it in the standard form of a geometric series $\sum_{n = 1}^{10}a\cdot r^{n - 1}$, we note that for the first deposit, when $n = 1$, the value is $300(1.055)^{10}$, for the second deposit when $n = 2$, the value is $300(1.055)^{9}$. We can rewrite the sum as $\sum_{n = 1}^{10}300\times1.055\times(1.055)^{n - 1}=\sum_{n = 1}^{10}316.5(1.055)^{n - 1}$. # Answer: $\sum_{n = 1}^{10}316.5(1.055)^{n - 1}$