# Lesson 26

Using the Sum

- Let’s calculate some totals.

### 26.1: Some Interesting Sums

Recall that for any geometric sequence starting at \(a\) with a common ratio \(r\), the sum \(s\) of the first \(n\) terms is given by \(s=a\frac{1-r^{n}}{1-r}\). Find the approximate sum of the first 50 terms of each sequence:

- \(\frac12\), \(\frac14\), \(\frac18\), \(\frac{1}{16}\), . . .
- 1, \(\frac12\), \(\frac14\), \(\frac18\), \(\frac{1}{16}\), . . .

### 26.2: That’s a lot of Houses

In 2012, about 71 thousand homes were sold in the United Kingdom. For the next 3 years, the number of homes sold increased by about 18% annually. Assuming the sales trend continues,

- How many homes were sold in 2013? In 2014?
- What information does the value of the expression \(71 \frac{(1-1.18^{11})}{(1-1.18)}\) tell us?
- Predict the total number of house sales from 2012 to 2017. Explain your reasoning.
- Do these predictions seem reasonable? Explain your reasoning.

Han and Lin each have a method to calculate \(3^5 + 3^6 + \ldots + 3^n\). Han says this is \(3^5 \left(1+ 3+ 3^2 + \ldots +3^{n-5}\right)\) and concludes that \(3^5+\ldots + 3^n = 3^5\frac{3^{n-4} -1}{3-1}\). Lin says that this is a difference of terms in 2 geometric sequences and can be written as \(\frac{3^{n+1}-1}{3-1} - \frac{3^5-1}{3-1}\). Do you agree with either Han or Lin? Explain your reasoning.

### 26.3: Back to Funding the Future

Let’s say you open a savings account with an interest rate of 5% per year compounded annually and that you plan on contributing the same amount to it at the start of every year.

- Predict how much you need to put into the account at the start of each year to have over $100,000 in it when you turn 70.
- Calculate how much the account would have after the deposit at the start of the 50th year if the amount invested each year were:
- $100
- $500
- $1,000
- $2,000

- Say you decide to invest $1,000 into the account at the start of each year at the same interest rate. How many years until the account reaches $100,000? How does the amount you invest into the account compare to the amount of interest earned by the account?

### Summary

Let’s say you plan to invest \$200 at the start of each year into an account that averages 3% interest compounded annually at the end of the year. How many years until the account has more than \$10,000? \$20,000?

We know that at the end of year 1 the amount in the account is \$206. At the end of year 2 the amount in the account is \$418.18 since \(200(1.03)^2 + 200(1.03)=418.18\). At the start of year 30, for example, that original \$200 has been compounded a total of 29 times while the last \$200 deposited has been compounded 0 times. Figuring out how much is in the account 30 years after the first deposit means adding up \(200(1.03)^{29} + 200(1.03)^{28} + . . . + 200(1.03) + 200\). We can use the formula for the sum of a geometric sequence, \(s=a\frac{(1-r^{n})}{(1-r)}\), to find the total amount in the account. The sequence starts at \(a=200\) and increases at a rate of \(r=1.03\) each year. After \(n\) years, the total \(s\) in the account is \(s=206\frac{(1-1.03^{n})}{(1-1.03)}\). Now we have a simpler expression to evaluate for different \(n\) values. It turns out that when \(n=31\), the account has about \$10,301 in it and when \(n=47\), it has about \$20,682 in it.

### Glossary Entries

**identity**An equation which is true for all values of the variables in it.