Lesson Video: Sigma Notation Mathematics

In this video, we will learn how to express a series in sigma notation and how to expand and evaluate series represented in sigma notation.

15:56

Video Transcript

In this video, we will learn how to express a series in sigma notation and how to expand and evaluate series represented in sigma notation. We will begin with some key definitions. Sigma notation is a concise and convenient way to represent long sums. For example, we often wish to sum a number of terms as shown, where there is an obvious pattern to the numbers involved. The first example is the sum of the first five positive intergers, and the second is the sum of the first six square numbers. More generally, if we consider the sequence of numbers 𝑒 sub one, 𝑒 sub two, 𝑒 sub three, and so on, all the way up to 𝑒 sub 𝑛, then we can write the sum of these numbers as 𝑒 sub one plus 𝑒 sub two plus 𝑒 sub three, and so on, all the way up to 𝑒 sub 𝑛.

If we let 𝑒 sub π‘Ÿ represent the general term, then this can be rewritten as the sum of 𝑒 sub π‘Ÿ from π‘Ÿ equals one to π‘Ÿ equals 𝑛. The symbol here is the Greek capital letter 𝛴, which corresponds to our letter S and therefore refers to the initial letter of the word sum. This expression means the sum of all the terms 𝑒 sub π‘Ÿ where π‘Ÿ takes values from one to 𝑛. It is important to note that our lower limit does not have to be equal to one. We can also write the sum of 𝑒 sub π‘Ÿ where π‘Ÿ takes values from π‘Ž to 𝑏. π‘Ž is the lower limit and 𝑏 is the upper limit. We will now consider some examples where we need to expand and evaluate a series given in sigma notation.

Expand and then evaluate the sum of two to the π‘Ÿth power minus 52, where π‘Ÿ takes values from one to four.

As π‘Ÿ can take values from one to four, we know there will be four terms in our series. These will be the values when π‘Ÿ is equal to one, two, three, and four. The Greek letter 𝛴 in the question means the sum. So we need to find the sum of these four terms. When π‘Ÿ is equal to one, we have two to the first power minus 52. When π‘Ÿ is equal to two, we have two squared minus 52. When π‘Ÿ is equal to three, we have two cubed minus 52. And finally, when π‘Ÿ is equal to four, we have two to the fourth power minus 52. We need to find the sum of these four terms.

Two minus 52 is equal to negative 50. As two squared is equal to four, the second term is negative 48. The third term is negative 44, and the fourth term is negative 36. The sum of these four negative values is negative 178. The sum of two to the π‘Ÿth power minus 52 from π‘Ÿ equals one to π‘Ÿ equals four is negative 178.

In our next question, we will look at a more complicated expression where the first term does not correspond to π‘Ÿ equals one.

Evaluate the sum of one-quarter multiplied by two to the power of π‘Ÿ minus one where π‘Ÿ takes values from four to nine.

The values four and nine are the lower and upper limits of π‘Ÿ, respectively. The Greek letter 𝛴 means the sum of. In this question, we need to find the sum of six terms when π‘Ÿ equals four, five, six, seven, eight, and nine. If we look at our expression, we notice that one-quarter is a constant. This means that we can rewrite our expression as shown: one-quarter multiplied by the sum of two to the power of π‘Ÿ minus one where π‘Ÿ takes values from four to nine. We now need to substitute each of the values of π‘Ÿ into our expression. Four minus one is equal to three. So we have two cubed. When π‘Ÿ is equal to five, we have two to the fourth power. Repeating this process, we have the six terms as shown.

We need to find the sum of these and then multiply our answer by one-quarter. Two cubed is equal to eight, two to the fourth power is equal to 16, and so on. As the six numbers inside our brackets sum to 504, we need to find one-quarter of 504. This is equal to 126. The sum of one-quarter multiplied by two to the power of π‘Ÿ minus one where π‘Ÿ takes values from four to nine is 126.

Our next example is a bit of a trick question. In all the questions we have seen so far, we have had a variable π‘Ÿ within the expression we’re trying to sum. However, in this case, we just have a constant, negative 25.

Evaluate the sum of negative 25 where π‘Ÿ takes values from two to six.

As the value we are trying to sum here is a constant, we might think that the answer is just negative 25. However, the sigma or sum notation tells us that our series will have five terms, when π‘Ÿ equals two, three, four, five, and six. Each of these terms will be equal to the constant negative 25. This means that we need to add negative 25, negative 25, negative 25, negative 25, and negative 25. This is the same as multiplying negative 25 by five, giving us negative 125.

In the remaining examples in this video, we will need to write a given series using sigma notation.

Express the series 54 multiplied by 12 plus 54 multiplied by 24 plus 54 multiplied by 36, and so on, all the way up to 54 multiplied by 240 in sigma notation.

In this question, we are given the first three terms of our series. Together with the 𝑛th or final term. We have been asked to express this series using sigma notation as shown. We need to find an expression for the general term 𝑒 sub π‘Ÿ and the number of terms in the series 𝑛. We notice that each of our terms is the product of two numbers. The first number is 54 in each of our terms. The second number corresponds to the multiples of 12. We have 12 times one, 12 times two, 12 times three, and so on. The final term is equal to 12 multiplied by 20. This means that the general term for our series is 54 multiplied by 12π‘Ÿ where π‘Ÿ is a variable going from one to 20.

54 multiplied by 12 is equal to 648. Therefore, 54 multiplied by 12π‘Ÿ is 648π‘Ÿ. Using sigma notation, we can therefore express the series as the sum of 648π‘Ÿ where π‘Ÿ takes values from one to 20.

We will now look at a more complicated series where our expression will involve exponents.

Express the series eight plus 32 plus 72 plus 128 and so on, all the way up to 512, in sigma notation.

In order to express any series using sigma notation, we need to write it in the form the sum of 𝑒 sub π‘Ÿ where π‘Ÿ takes values from one to 𝑛. We need to work out the expression for 𝑒 sub π‘Ÿ, which is the general term of the series. At first glance, there doesn’t appear to be an obvious link between each of the terms in our series. However, they do all have a common factor of eight. Eight is eight multiplied by one, 32 is eight multiplied by four, 72 is eight multiplied by nine, and so on. One, four, nine, 16, and so on are the square numbers. This means that the general term of our series is eight multiplied by π‘Ÿ squared. As eight squared is equal to 64, our lower and upper limits are equal to one and eight.

The given series expressed using sigma notation is the sum of eight π‘Ÿ squared, where π‘Ÿ takes values from one to eight. We could check this answer by substituting in each of our values of π‘Ÿ which would obtain the terms eight, 32, 72, and so on.

We will now consider one final example.

Express the series 496 minus 497 plus 498 minus 499 and so on, all the way up to minus 531, in sigma notation.

We notice here that the terms in our series alternate from positive to negative. A series can be written using sigma notation as shown: the sum of the general term 𝑒 sub π‘Ÿ where π‘Ÿ takes values from one to 𝑛. In this question, we need to find an expression for this general term 𝑒 sub π‘Ÿ as well as the value of 𝑛. Let’s begin by considering the sequence of positive integers 496, 497, 498, and so on, all the way up to 531. We can rewrite each of these terms as the constant 495 plus some integer. 496 is 495 plus one, 497 is 495 plus two, and so on, all the way up to 531 which is equal to 495 plus 36. This suggests that we have 36 terms in our sequence.

If we were looking to write this series in sigma notation, we would have the sum of 495 plus π‘Ÿ where π‘Ÿ takes values from one to 36. The series in this question is slightly more complicated, though, as we want the second, fourth, sixth, and every other even term to be negative. We can make any positive integer negative by multiplying by negative one. If we multiplied each of our terms by negative one to the power of π‘Ÿ, the first, third, fifth, and all our terms would be negative. As we want the even terms to be negative, we need to multiply by negative one to the power of π‘Ÿ plus one. The series 496 minus 497 plus 498 and so on can be written in sigma notation as negative one to the power of π‘Ÿ plus one multiplied by π‘Ÿ plus 495, where π‘Ÿ takes values from one to 36.

We will now summarize the key points from this video. The sum 𝑒 sub one plus 𝑒 sub two plus 𝑒 sub three and so on, all the way up to 𝑒 sub 𝑛, is written in sigma notation as the sum of 𝑒 sub π‘Ÿ, where π‘Ÿ takes values from one to 𝑛. In this video, we saw that we could expand the series given the sigma notation, as well as write the series in sigma notation. 𝑒 sub π‘Ÿ is the general term of our series. Whilst we normally take this sum of terms from π‘Ÿ equals one to π‘Ÿ equals 𝑛, we can also sum from π‘Ž to 𝑏, where π‘Ž is the lower limit and 𝑏 is the upper limit.

We also saw in two of our examples some key rules when dealing with sigma notation. If 𝐢 is a constant, the sum of 𝐢 from π‘Ÿ equals one to 𝑛 is equal to 𝑛 multiplied by the constant 𝐢. Also, the sum of some constant 𝐢 multiplied by the variable π‘Ÿ from π‘Ÿ equals one to π‘Ÿ equals 𝑛 is equal to the constant 𝐢 multiplied by the sum of π‘Ÿ from π‘Ÿ equals one to 𝑛. We can factor out the constant, calculate the sum separately, and then multiply this answer by the constant 𝐢.

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