Some Special Series (Part 2)

Quick Navigation Geometric Series Exercises

Geometric Series

A geometric series is a series formed by adding together the terms of a geometric sequence.

In the first part of this section, we looked at formulas that you could use to find the sum of arithmetic series. There's a similar formula for the sum of a geometric series but it doesn't have the kind of neat origin that the arithmetic formula does. For this one, we're going to have to fall back on some algebra. This is probably one of the most complicated derivations in the entire book so if you don't get it at first, don't let that frustrate you. You can always jump ahead to the final formula at the end of the calculations.

First, assume that our geometric series has the formula $ar^i$ where i goes from 0 to n and a is the first term. That means we're looking for

$$\sum_{i=0}^n ar^i$$

$$S=\sum_{i=0}^n ar^i$$	Let S equal the sum. Now that we have an equation, we can manipulate it to try to come up with a formula.
$$rS=r\sum_{i=0}^n ar^i$$	I'm going to start by multiplying both sides of the equation by r.
$$rS=r\sum_{i=0}^n ar^i=\sum_{i=0}^n r\cdot ar^i=\sum_{i=0}^n ar^{i+1}$$	On this step, I used the Distributive Property to bring the r inside the sum and then I used the exponent rules to simplify r · rⁱ to r^{i + 1}.
$$rS=\sum_{j=1}^{n+1} ar^{j}$$	That step was a little tricky. I wanted to get rid of the +1 in the exponent so I did it by changing the counter from "0 to n" to "1 to n + 1". If you try substituting some numbers for j, you'll see that I get the same series. Both of them start at ar and end with ar^{n + 1}. (Be sure that you're comparing this series to the one in the previous step, not the original.)
$$rS=\sum_{j=1}^{n} ar^{j} + ar^{n+1}$$	For this step, I took the n + 1 term out of the sigma part and explicitly added it to the end.
$$a + rS=\sum_{j=0}^{n} ar^{j} + ar^{n+1}$$	This step is also a little tricky. I changed the starting point of the series from 1 to 0 which had the effect of adding ar⁰ or just a to the right side. To balance that out, I also added a to the left hand side.
$$rS=\sum_{j=0}^{n} ar^{j} + ar^{n+1} - a$$	I moved the a over to the right side of the equation by subtracting it from both sides.
$$rS = S + ar^{n-1} - a$$	Notice that the $\sum_{i=0}^n ar^i$ on the right side of the last line is just our original series. (The counter is different, j instead of i, but the terms in the sum are the same which is what really matters.) Because the original series is equal to S, I substituted that into the previous equation for $\sum_{j=0}^n ar^j$.
$$rS - S = ar^{n+1} - a$$ $$S(r - 1) = ar^{n+1} - a$$ $$S = \frac{ar^{n+1}-a}{r-1}$$	Now, I just took the previous equation and solved it for S. Because S is the original series, this gives us our final formula.

Geometric Series

$\sum_{i=0}^n ar^i =\frac{ar^{n + 1} - a}{r - 1}$ assuming that $r \ne 1$.

Example 1 - Geometric Series

Find the sum of the series 2 + 6 + 18 + ... + 4374.

The key to this kind of question is first recognizing that it's a geometric series. Looking at the first two terms, it looks like we're starting at 2 and then multiplying each term by r = 3 to get the next one. That gives us two of the things that we need to know to use our formula

a = 2r = 3

The one thing that we still need to find is the value of n. We can do that using a little algebra. We know that the formula for our series looks like 2 · 3ⁱ and we need to find the value of i that gives our final value, 4374. In other words, we need to know the value of n that makes

$$2 \cdot 3^n = 4374$$

Solving that equation gives us

$$2 \cdot 3^{n} = 4374$$ $$3^{n} = 2187$$ $$\log_{3}3^{n} = \log_{3}2187$$ $$n = \log_{3}2187$$ $$n = \frac{\log(2187)}{\log(3)} = 7$$

It's an important sign that we ended up with an integer. If your calculations gave you a decimal value, then there was a problem somewhere that you needed to go back and fix.

Now that we have values for a, r and n, we can get the sum by using our formula.

$$2 + 6 + 18 + ... + 4374 = \sum_{i=0}^7 2 \cdot 3^i=\frac{2 \cdot 3^{7+1} - 2}{3 - 1} = 6560$$

Example 2 - Geometric Series

Find the sum of the series 4 - .4 + .04 - .004 + .0004.

In this case, we're given the entire series so we can see right away that a = 4 and n = 4. If you divide several of the consecutive terms, you'll see that

$$\frac{4}{-.4)} = -.1$$ $$\frac{-.4}{.04} = -.1$$ $$\frac{.04}{-.004} = -.1$$

. . .

This tells us that the series is a geometric series with r = -.1. Don't let the negative value bother you. Our formula works for any value of r except for 1. Substituting those values into the formula gives us

$$4 - .4 + .04 - .004 + .0004 = \sum_{i=0}^4 4 \cdot (-.1)^i=\frac{4 \cdot (-.1)^{4 + 1} - 4}{-.1 - 1} = 3.6364$$

Exercises

= show solution = get hint

Exercises

1 + 2 + 3 + ... + 999 + 1000
80 + 82 + ... + 104 + 105
1 + 2 + 4 + ... + 64 + 128
1 + 2 + 3 + ... + 755 + 756
3 + .3 + .03 + ... + .000003
6 + 30 + 150 + ... + 18750
3 - 6 + 12 + ... + 192 - 384
1 + 2 + 3 + ... + 75 + 76
101 + 102 + ... + 199 + 200
10 + 2 + .4 + ... + .00064
99 + 100 + ... + 1005 + 1006
1 + 1.1 + 1.21 + ... + 1.771561

Analysis

In the formula for the sum of the first n integers, I talked about n / 2 being the number of pairs in the list but what happens if n is odd? Then n / 2 is a decimal but the sum of a list of integers has to come out to an integer, i.e. it can't be a decimal. Give an argument for why n(n + 1) / 2 will always be an integer. (Hint: If n is odd then what is n + 1?)
Find the sum of the multiples of 3 from 3 to 243, i.e. 3 + 6 + 9 + ... + 240 + 243.
Based on the calculations that we did in Example 3, propose a formula for the sum of the even integers from 2 to n, assuming that n is even.
Calculate the sum of the odd numbers from 1 to 81. (Hint: In Example 3, we used 2i for 'any even number'. In this calculation, try using 2i + 1 for 'any odd number'. Try substituting some values into that expression for i to convince yourself that it always gives you an odd number.)