Lesson Video: The General Term of a Sequence Mathematics

In this video, we will learn how to use the general term or a recursive formula of a sequence to work out terms in the sequence.

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Video Transcript

In this video, we will learn how to use the general term or a recursive formula of a sequence to work out the terms in a sequence. We begin by recalling that a sequence is an ordered list of numbers, each of which is called a term, for example, the even numbers two, four, six, eight, 10, and so on. When dealing with sequences, we can usually find the next term by spotting a general rule or pattern. In this video, we will extend this to look at the general term and how we can use this to find any term in a sequence.

The general term of a sequence, sometimes called the 𝑛th term and written π‘Ž sub 𝑛, is an algebraic expression that relates the term to its position number in the sequence. Let’s consider the general term π‘Ž sub 𝑛 equal to three 𝑛 plus four. One way of calculating the numbers in this sequence is to set up a table as shown. We will begin by calculating the first three terms, when 𝑛 is equal to one, two, and three. When 𝑛 is equal to one, we have three multiplied by one plus four. This is equal to seven, so the first term in our sequence is seven.

The second term can be calculated by multiplying two by three and then adding four. This is equal to 10. Following the same method, we see that the third term is 13. This means that π‘Ž sub one is equal to seven, π‘Ž sub two is equal to 10, and π‘Ž sub three is equal to 13. The sequence with general term π‘Ž sub 𝑛 equals three 𝑛 plus four is seven, 10, 13, and so on.

Let’s now consider how we could calculate the eighth term in this sequence. Substituting 𝑛 equals eight into our expression, we have three multiplied by eight plus four. This is equal to 28. π‘Ž sub eight, the eighth term in the sequence, is therefore equal to 28. This method can be summarized as follows. If the general term of a sequence contains an expression in 𝑛, we substitute the term number in place of 𝑛 to find a specific term in the sequence. For example, to find the 20th term, we substitute 𝑛 equals 20 into our expression.

In our first example, we will use this method to find the first five terms of a sequence.

Find the first five terms of a sequence whose 𝑛th term is given by π‘Ž sub 𝑛 equals 𝑛 squared minus 14, where 𝑛 is greater than or equal to one.

In order to calculate the first five terms of our sequence, we need to substitute 𝑛 equals one, two, three, four, and five into our expression for π‘Ž sub 𝑛. The first term is when 𝑛 equals one. We denote this as π‘Ž sub one, and it is equal to one squared minus 14. We know that one squared is equal to one, and subtracting 14 from this gives us negative 13. The first term of our sequence is therefore negative 13. To calculate the second term, we substitute 𝑛 equals two. This is equal to two squared minus 14. And as two squared is equal to four, this is equal to negative 10.

We can repeat this process to calculate the third, fourth, and fifth terms. When 𝑛 equals three, π‘Ž sub three is equal to negative five. When 𝑛 equals four, four squared minus 14 equals two. And finally, the fifth term π‘Ž sub five is equal to five squared minus 14, which equals 11. The first five terms of a sequence whose 𝑛th term is given by π‘Ž sub 𝑛 equals 𝑛 squared minus 14 are negative 13, negative 10, negative five, two, and 11.

In our next example, we need to calculate a specific term of a sequence.

Find the seventh term of the sequence π‘Ž sub 𝑛 is equal to 𝑛 cubed minus 14.

The seventh term of a sequence with a general term π‘Ž sub 𝑛 will be denoted π‘Ž sub seven. We can calculate this by substituting 𝑛 equals seven into our expression for the general term. π‘Ž sub seven is equal to seven cubed minus 14. We know that seven multiplied by seven or seven squared is equal to 49. This means that we can calculate seven cubed by multiplying 49 by seven. As 40 multiplied by seven is 280 and nine multiplied by seven is 63, we can find the sum of these to calculate 49 multiplied by seven, which is 343. The seventh term of the sequence is equal to 343 minus 14. This is equal to 329. We can use this method to calculate any term in a sequence when given an expression for the general term in terms of 𝑛. For example, to calculate the 20th term, we would substitute 𝑛 equals 20 into the expression.

Before looking at our next example, we will consider what it means to have a recursive formula for our general term. A sequence can be defined by giving one general term of the sequence as an expression using other terms of the sequence. If this relationship between terms occurs throughout the sequence, it is called a recurrence relation. In its simplest form, this allows us to calculate the next term of a sequence given the previous one. Let’s consider the general term with recursive formula π‘Ž sub 𝑛 is equal to two multiplied by π‘Ž sub 𝑛 minus one plus five, where 𝑛 is greater than or equal to two and π‘Ž sub one is equal to four.

We see that the expression here contains the term π‘Ž sub 𝑛 minus one. This is the term immediately prior π‘Ž sub 𝑛. To calculate any term in this sequence, we multiply the previous term by two and then add five. We can once again demonstrate this in a table. We are told that π‘Ž sub one is equal to four. Therefore, when 𝑛 is equal to one, π‘Ž sub 𝑛 is four. To calculate the second term in the sequence, we multiply the first term four by two and then add five. This gives us 13.

We can then repeat this process to calculate the third term. We multiply the second term by two and add five. Two multiplied by 13 plus five is equal to 31. Finally, in order to calculate the fourth term, we multiply the third term by two and add five once again. This gives us 67. This sequence will carry on indefinitely. The first four terms of the sequence with general term π‘Ž sub 𝑛 is equal to two multiplied by π‘Ž sub 𝑛 minus one plus five are four, 13, 31, and 67. We will now summarize how we can use the recurrence relation or recursive formula.

If the general term of a sequence contains an expression in π‘Ž sub 𝑛 minus one, we substitute the preceding term in place of π‘Ž sub 𝑛 minus one to find any term. If, however, the general term contains an expression for π‘Ž sub 𝑛 plus one in π‘Ž sub 𝑛, we substitute the preceding term in place of π‘Ž sub 𝑛 to find the value of π‘Ž sub 𝑛 plus one. One example of a sequence that can be defined using a recursive formula is the Fibonacci sequence. This contains the numbers one, one, two, three, five, eight, 13, and so on. To calculate the next term in the Fibonacci sequence, we add the two preceding numbers. For example, one plus one is equal to two, one plus two equals three, two plus three equals five, and so on. This is a really interesting sequence that has many applications in nature that you may like to research further.

We will now look at a specific example involving a recursive formula.

Find the first five terms of the sequence π‘Ž sub 𝑛, given π‘Ž sub 𝑛 plus one is equal to negative one to the 𝑛th power divided by nine multiplied by π‘Ž sub 𝑛, where 𝑛 is greater than or equal to one and π‘Ž sub one equals negative 11.

In this question, we are given the value of the first term of our sequence π‘Ž sub one, which is equal to negative 11. We are asked to find the first five terms. So we therefore need to calculate the second, third, fourth, and fifth terms of the sequence. These are denoted π‘Ž sub two, π‘Ž sub three, π‘Ž sub four, and π‘Ž sub five.

The formula we are given is an example of a recursive formula, where π‘Ž sub 𝑛 plus one is given in terms of π‘Ž sub 𝑛. We can therefore calculate any term of the sequence by substituting in the previous term. The second term π‘Ž sub two is therefore equal to negative one to the power of one divided by nine multiplied by π‘Ž sub one. We know that π‘Ž sub one is equal to negative 11. This expression is equal to negative one over negative 99, which in turn simplifies to one over 99. This is the second term of the sequence, and we can use this to calculate the third.

Using our recursive formula once again, π‘Ž sub three is equal to negative one squared over nine multiplied by π‘Ž sub two. We know that π‘Ž sub two is one over 99. On the denominator, we have nine multiplied by one over 99, which is the same as nine over 99. Dividing the numerator and denominator by nine, this simplifies to one over 11 or one eleventh. As negative one squared is equal to one, π‘Ž sub three is equal to one divided by one eleventh. We know that dividing by a fraction is the same as multiplying by the reciprocal of this fraction. One divided by one eleventh is the same as one multiplied by 11 over one. This is equal to 11. π‘Ž sub three, the third term in our sequence, is 11.

Next, we need to calculate the fourth term π‘Ž sub four. This is equal to negative one cubed divided by nine multiplied by π‘Ž sub three. And we have just calculated that π‘Ž sub three equals 11. The fourth term of the sequence is therefore equal to negative one over 99, recalling that cubing a negative number gives a negative answer. Finally, we have π‘Ž sub five is equal to negative one to the fourth power divided by nine multiplied by π‘Ž sub four. We can substitute negative one over 99 for π‘Ž sub four. And this in turn simplifies to one over negative one eleventh. Using a similar method that we used to calculate π‘Ž sub three, we have π‘Ž sub five is equal to negative 11. The first five terms of the sequence are negative 11, one over 99, 11, negative one over 99, and negative 11.

We notice that the first term is equal to the fifth term. This means that we have a periodic sequence where the first four terms are repeated. π‘Ž sub one is equal to π‘Ž sub five, which is equal to π‘Ž sub nine and so on. Likewise, the second, sixth, and 10th terms are equal. The same is true for the third, seventh, and 11th terms, together with the fourth, eighth, and 12th terms.

In our final example, we will determine which of the expressions corresponds to the given sequence.

Which of the following is an expression for the general term of the sequence 52, 84, 116, 148? Is it (A) 52 plus 30 multiplied by 𝑛 minus one? (B) 52 plus 32 multiplied by 𝑛 minus one. (C) 52 plus 32 multiplied by 𝑛 plus one. (D) 84 plus 32 multiplied by 𝑛 minus one. Or (E) 84 plus 30 multiplied by 𝑛 plus one.

We are told in the question that the first four terms of our sequence are 52, 84, 116, and 148. These correspond to the integer values of 𝑛, one, two, three, and four. In order to work out which of the expressions corresponds to this sequence, we can substitute these values into each expression in turn. Let’s begin with 𝑛 equals one. In option (A), we have 52 plus 30 multiplied by one minus one. As one minus one equals zero, this is equal to 52. This means that expression (A) does satisfy the first term of our sequence. This is also true of option (B). 52 plus 32 multiplied by one minus one is equal to 52. Options (C), (D), and (E) give values of 116, 84, and 144 when we substitute 𝑛 equals one. This means that these expressions do not have a first term equal to 52. And we can therefore rule out these options.

We will now substitute 𝑛 equals two into the expressions (A) and (B). In option (A), we have 52 plus 30 multiplied by two minus one, which is equal to 82. In option (B), we obtain the answer 84. Since the second term of our sequence is 84, we can rule out option (A). Whilst it appears that option (B) is the correct answer, it is worth checking that this is the correct expression for 𝑛 equals three and 𝑛 equals four. When 𝑛 is equal to three, the expression 52 plus 32 multiplied by 𝑛 minus one gives us 116. And when 𝑛 is equal to four, the expression gives us 148. These do correspond to the four terms of our sequence, and the correct answer is therefore option (B).

We will now summarize the key points from this video. The general term of a sequence is an expression that relates the term to its position number or the term that precedes it. If the general term contains an expression in 𝑛, then we substitute the term number in place of 𝑛. However, if the general term contains an expression in π‘Ž sub 𝑛 minus one, we substitute the preceding term in place of π‘Ž sub 𝑛 minus one.

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