An arithmetic series is the sum of an arithmetic sequence. A geometric series is the sum of a geometric sequence. There are other types of series, but you're unlikely to work with them until you're in calculus. For now, you'll probably just work with these two.

For reasons that will be explained in calculus, you can only take the partial sum of an arithmetic sequence. The "partial" sum is the sum of a limited (that is to say, finite) number of terms, like the first ten terms, or the fifth through the hundredth terms.

The formula for the first n terms of an arithmetic sequence, starting with n = 1, is:

The sum is, in effect, n times the "average" of the first and last terms. This sum of the first n terms is called "the n-th partial sum". (By the way: The above summation formula can be proved usinginduction.)

Find the 35th partial sum of an = (1/2)n + 1

The 35th partial sum of this sequence is the sum of the first thirty-five terms. The first few terms of the sequence are:

a1 = (1/2)(1) + 1 = 3/2 a2 = (1/2)(2) + 1 = 2 a3 = (1/2)(3) + 1 = 5/2

The terms have a common difference d = 1/2, so this is indeed an arithmetic sequence. The last term in the partial sum will be a35 = a1 + (35 – 1)(d) = 3/2 + (34)(1/2) = 37/2. Then, plugging into the formula, the 35th partial sum is:

(n/2)(a1 + an) = (35/2)(3/2 + 37/2) = (35/2)(40/2) = 350

Find the value of the following summation

From the formula ("2n – 5") for the n-th term, I can see that each term will be two units larger than the previous term. (Plug in values for n if you're not sure about this.) So this is indeed an arithmetical sum. But this summation starts at n = 15, not at n = 1, and the summation formula applies to sums starting at n = 1. So how can I work with this summation?

The quickest way to find the value of this sum is to find the 14th and 47th partial sums, and then subtract the 14th from the 47th. By doing this subtraction, I'll be left with the value of the sum of the 15th through 47th terms. The first term is a1 = 2(1) – 5 = –3. The other necessary terms are a14 = 2(14) – 5 = 23 and a47= 2(47) – 5 = 89.

Subtracting, I get:

Then the solution is: Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved

Find the value of n for which the following equation is true:

I know that the first term is a1 = 0.25(1) + 2 = 2.25. I can see from the formula that each term will be 0.25 units bigger than the previous term, so this is an arithmetical series. Then the summation formula for arithmetical series gives me:

(n/2)(2.25 + [0.25n + 2]) = 21 n(2.25 + 0.25n + 2) = 42 n(0.25n + 4.25) = 42 0.25n2 + 4.25n – 42 = 0 n2 + 17n – 168 = 0 (n + 24)(n – 7) = 0

Solving the quadratic, I get that n = –24 (which won't work in this context) or n = 7.

n = 7

You could do the above exercise by adding terms until you get to the required total of "21". But your instructor could easily give you a summation that requires, say, eighty-six terms before you get the right total. So make sure you can do the computations from the formula.

Find the sum of 1 + 5 + 9 + ... + 49 + 53.

Checking the terms, I can see that this is indeed an arithmetic series: 5 – 1 = 4, 9 – 5 = 4, 53 – 49 = 4. I've got the first and last terms, but how many terms are there in total?

I have the n-th term formula, "an = a1 + (n – 1)d", and I have a1 = 1 and d = 4. Plugging these into the formula, I can figure out how many terms there are:

an = a1 + (n – 1)d 53 = 1 + (n – 1)(4) 53 = 1 + 4n – 4 53 = 4n – 3 56 = 4n 14 = n

So there are 14 terms in this series. Now I have all the information I need:

1 + 5 + 9 + ... + 49 + 53 = (14/2)(1 + 53) = (7)(54) = 378

You can take the sum of a finite number of terms of a geometric sequence. And, for reasons you'll study in calculus, you can take the sum of an infinite geometric sequence, but only in the special circumstance that the common ratio r is between –1 and 1; that is, you have to have | r | < 1.

For a geometric sequence with first term a1 = a and common ratio r, the sum of the first n terms is given by: Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved

Note: Your book may have a slightly different form of the partial-sum formula above. For instance, the "a" may be multiplied through the numerator, the factors in the fraction might be reversed, or the summation may start at i = 0 and have a power of n + 1 on the numerator. All of these forms are equivalent, and the formulation above may be derived from polynomial long division.

In the special case that | r | < 1, the infinite sum exists and has the following value:

Evaluate the following:

The first few terms are –6, 12, –24, so this is a geometric series with common ratio r = –2. (I can also tell that this must be a geometric series because of the form given for each term: as the index increases, each term will be multiplied by an additional factor of –2.) The first term of the sequence is a = –6. Plugging into the summation formula, I get:

So the value of the summation is 2 097 150

Evaluate S10 for 250, 100, 40, 16,....

I need to find the sum of the first ten terms. The first term is a = 250. Dividing pairs of terms, I get 100 ÷ 250 = 2/5, 40 ÷ 100 = 2/5, etc, so the terms being added form a geometric sequence with common ratio r = 2/5. When I plug in the values of the first term and the common ratio, the summation formula gives me:

Find an if S4 = 26/27 and r = 1/3.

They've given me the sum of the first four terms, S4, and the value of the common ratio r. Since there is a common ratio, I know this must be a geometric series. Plugging into the geometric-series-sum formula, I get:

Multiplying on both sides by 27/40 to solve for the first term a = a1, I get:

Then:

Show, by use of a geometric series, that 0.3333... is equal to 1/3.

There's a trick to this. I first have to break the repeating decimal into separate terms:

0.333... = 0.3 + 0.03 + 0.003 + 0.0003 + ...

This shows the repeating pattern of the non-terminating (never-ending) decimal explicitly: For each term, I have a decimal point, followed by a steadily-increasing number of zeroes, and then ending with a "3". This can be written in fractional form, and then converted into geometric-series form:

Then 0.333... is an infinite geometric series with a = 3/10 and r = 1/10. Since | r | < 1, I can use the formula for summing infinite geometric series:

Using the summation formula to show that the geometric series "expansion" of 0.333... has a value of one-third is the "showing" that the exercise asked for. You can use this method to convert any repeating decimal to its fractional form:

By use of a geometric series, convert 1.363636... to fractional form.

First I'll break this into its constituent parts, so I can find the pattern:

1.363636.. = 1 + 0.36 + 0.0036 + 0.000036 + ...

There are two digits that repeat, so the fractions are a little bit different. But this is still a geometric series:

Then this is the leading "1", plus a geometric series having a = 9/25 and r = 1/100. Then the sum is:

Note: This technique can also be used to convert any repeating decimal into fractional form, and also can be used to prove that 0.999... = 1.

A sequence is a list of numbers. A series is created by adding terms in the sequence. This lesson assumes that you know about arithmetic sequences, how to find the common difference and how to find an explicit formula. You may want to review the basics of arithmetic sequences or finding formulas. 

There are two ways to indicate that you are adding terms in a sequence. One is by using summation notation and one is by using subscript notation similar to how we write explicit forms of sequences. 

Let’s look at each of these methods separately.

1. First, summation notation. In summation notation, you are given an expression and told how many terms you are to add up.

 tells you to use the values of n = 1, n = 2, and n = 3 in the expression 2n + 1. 

Once those values are substituted in, you add them all up.

Summary

In general, summation notation looks like   wherean is an expression with n as the variable andk tells you how many terms to add.

Let's Practice summation

notation: Evaluate   This problem is asking us to add up five terms. So use n = 1, 2, 3, 4, and 5 in the expression -3n +5.

2. The series created by adding the first five terms of the sequence in the previous example can be found using another method called subscript notation.

Summary

To find the sum of terms in an arithmetic sequence, use the following formula.

In this formula:Sn is the sum of the first n terms in a sequencen is the number of terms you are adding upa1 is the first term of the sequencean is the nth term of the sequenceThe nth term is found by using the explicit formula for the sequence. So if you are not given the explicit formula, you will need to find it before you can find the sum. 

Let's Practice subscript notation:

i. Find   for the sequence  . 

The formula says we need to know n, the first term, and the nth term.n = 10 since we are asked to find the sum of the first 10 terms

Substituting these values into the equation gives

ii.Find the sum of the first 21 terms of the sequence 3, 7, 11, 15, . . . 

The formula says we need to know n, the first term, and the nth term.n = 21 since we are asked to find the sum of the first 21 terms. 

To find a1 and a21 we will need the explicit formula.

So now we can find

We now have everything we need to find the sum of the first 21 terms.