Lesson Explainer: Sequence Formulas Mathematics

In this explainer, we will learn how to find the general term or a recursive formula for a sequence and how to use them to work out terms in the sequence.

A sequence can be described by a general term using a position index, 𝑛. For example, we can write the sequence of even numbers, 2,4,6,8,, in terms of index values 𝑛1 as

Index, 𝑛1234
Term2468

Any 𝑛th term in the sequence is written as 𝑇. Thus, the first term can be written as 𝑇, the second term as 𝑇, and so forth.

When studying sequences, we aim to find a general rule for the sequence such that if we wanted to find a term with a larger index, we can do so without listing all the terms up to that specific one. For the given sequence, 2,4,6,8,, we can recognize that each sequence term is double its index.

We can write this general rule more mathematically, stating that, for any 𝑛th term 𝑇, where 𝑛1, 𝑇=2𝑛.

Thus, to calculate a specific term, for instance, the 23rd term, we would substitute the value 𝑛=23 into the general term 𝑇=2𝑛 to find the value of 𝑇 as 𝑇=2×23=46.

In the following example, we will see how we can find the first five terms of a sequence given its 𝑛th term.

Example 1: Finding the Terms of a Sequence given the General Term

Find the first five terms of the sequence whose general term is given by 𝑇=𝑛(𝑛34), where 𝑛1.

Answer

In this question, we are given the general, or 𝑛th, term of a sequence for an index 𝑛. To find any term in the sequence, we substitute that value for 𝑛 into the general term.

Therefore, for the first term, 𝑇, we substitute 𝑛=1 into 𝑇=𝑛(𝑛34) and simplify, giving 𝑇=1(134)=1(33)=33.

We can continue to find the second term, 𝑇, by substituting 𝑛=2 into the general term, giving 𝑇=2(234)=2(32)=64.

The third term, 𝑇, is calculated as 𝑇=3(334)=3(31)=93.

The fourth term, 𝑇, is calculated as 𝑇=4(434)=4(30)=120.

And the fifth term, 𝑇, is calculated as 𝑇=5(534)=5(29)=145.

Hence, we can list the first five terms of the sequence as 33,64,93,120,145.

In the previous example, we had a general term that related the terms of the sequence directly to their index value. For some sequences, however, the 𝑛th term of the sequence can be given related to the term immediately preceding it.

Let us take the example of the sequence 1,4,7,10,. We can compare the values of the index 𝑛1 to the values of 3𝑛.

Index, 𝑛1234
3𝑛36912

The values of 3𝑛 do not give us the same terms of the sequence, but we can obtain these sequence terms by subtracting 2 from each of the values of 3𝑛.

Index, 𝑛1234
3𝑛36912
3𝑛214710

Therefore, for 𝑛1, we could write the 𝑛th term of the sequence 1,4,7,10, as 𝑇=3𝑛2.

Alternatively, we could also note that the pattern between terms is to add 3 to the previous term.

Thus, any term 𝑇 is equal to the previous term plus 3. Using index notation, we can write the term before 𝑇 as 𝑇.

Therefore, we could write that this sequence has an 𝑛th term of 𝑇=𝑇+3. When we write a formula in this way, we also need to indicate the value of the first term, 𝑇. Therefore, the 𝑛th term can be given as 𝑇=1;𝑇=𝑇+3,𝑛2.

Note that we can include the limits on the value of the index, 𝑛2, as this formula applies only for values of 𝑇 greater than or equal to 𝑇.

The formula for a sequence given in this way is called a recursive formula.

Definition: Recursive Formula for a Sequence

A recursive formula is a formula in which the terms of a sequence are defined using one or more of the previous terms.

We may see a recursive formula given for 𝑇 or 𝑇. For example, we defined the recursive formula for the sequence 1,4,7,10, as 𝑇=1;𝑇=𝑇+3,𝑛2, but we could have also defined this as 𝑇=1;𝑇=𝑇+3,𝑛1.

Both formulas demonstrate recursion of the term before (either 𝑇 and 𝑇, or 𝑇 and 𝑇). The difference will be in the first value of the index, 𝑛, that can be used.

In the next example, we will see how we can find a specific term in a sequence, given a recursive formula.

Example 2: Finding Terms of a Sequence given a Recursive Formula

If (𝑇) is a sequence defined as 𝑇=11 and 𝑇=𝑇3, where 𝑛1, then the fourth term equals .

  1. 2
  2. 4
  3. 5
  4. 8

Answer

In this question, we are given a recursive formula for a sequence. In a recursive formula, the terms of the sequence are defined using one or more of the previous terms. Thus, to find a specific term in the sequence, we cannot simply substitute in a value for the index and immediately get the value of the term; rather, we need to find the preceding term(s). This means we may need to apply the recursive formula several times.

The recursive formula here is given as 𝑇=𝑇3. This means that any term, 𝑇, in the sequence is found by the previous term, 𝑇, minus 3.

The fourth term, 𝑇, will be found by 𝑇=𝑇3. The third term, 𝑇, can be found using the second term, and the second term can be found by using the first term. Therefore, we can begin at the start of the sequence and compute each term.

To apply this recursive formula, we need the first term of the sequence, which is given here as 𝑇=11.

The second term of the sequence can be found by substituting 𝑛=1 and 𝑇=11 into 𝑇=𝑇3 to give 𝑇=𝑇3=113=8.

To calculate the third term of the sequence, we substitute the second term, 𝑇=8, and 𝑛=2 to give 𝑇=𝑇3=83=5.

Finally, we can compute the required term of the sequence, 𝑇, as 𝑇=𝑇3=53=2.

Hence, we can give the answer that the fourth term of the sequence is that given in option A, 2.

Next, we will look at some properties of sequences, and how to determine if a sequence is increasing or decreasing.

Let us take the example of the sequence of square numbers that has the 𝑛th term 𝑇=𝑛, for 𝑛1. The first 4 terms of the sequence can be found as 1,4,9,16,.

Each term in the sequence is always greater than the previous term; therefore, this is an increasing sequence.

If, instead, we examined the sequence 𝑇=1𝑛, for 𝑛1, and calculated the first 4 terms in the sequence as 1,12,13,14,, then, as each term is smaller than the previous term, this is a decreasing sequence.

We can formalize these definitions below.

Definition: Monotonic Increasing and Decreasing Sequences

A sequence of real numbers (𝑇) is said to be increasing if 𝑇>𝑇 for all 𝑛.

A sequence of real numbers (𝑇) is said to be decreasing if 𝑇<𝑇 for all 𝑛.

A sequence of real numbers (𝑇) is said to be constant if 𝑇=𝑇 for all 𝑛.

If a sequence is increasing, decreasing, or constant, it is called a monotonic sequence.

In the next example, we will see how to apply these rules to identify the properties of a sequence.

Example 3: Determining the Nature of a Sequence

Is the sequence 𝑇=(1)11𝑛22 increasing, decreasing, or neither?

Answer

In this question, we are given the 𝑛th term of a sequence and asked to determine the properties of this sequence. We recall that a sequence (𝑇) is said to be increasing if 𝑇>𝑇 for all 𝑛. Every term in the sequence must be greater than the previous term.

Conversely, a sequence is decreasing if each term in it is less than the previous term, such that 𝑇<𝑇.

We can find the first few terms in the sequence in order to help determine its properties.

For the first term, we can substitute the value 𝑛=1 into 𝑇=(1)11𝑛22 to give 𝑇=(1)11×122=11122=24311.

The second term, 𝑇, can be found by substituting 𝑛=2 into the 𝑛th term, giving 𝑇=(1)11×222=12222=48322.

The third and fourth terms can be calculated as 𝑇=(1)11×322=13322=72733 and 𝑇=(1)11×422=14422=96744.

We can write first 4 terms in the sequence, fractionally, as 24311,48322,72733,96744,.

It may be helpful to calculate the decimal approximation of each sequence term, rounded to 2 decimal places, as 22.09,21.95,22.03,21.98,.

We note that the second term is less than the first term, so the sequence is not increasing, and the third term is more than the second, so the sequence is not decreasing.

Therefore, we can give the answer that 𝑇.isneitherincreasingnordecreasing

Thus far, we have identified the properties, and determined specific terms of a sequence. We will now see how we can find the general term of a sequence, given the first terms of that sequence.

Example 4: Finding the General Term of a Sequence

Find, in terms of 𝑛, the general term of the sequence (18,72,162,288,).

  1. 18𝑛
  2. 18𝑛
  3. 18𝑛
  4. 19𝑛1
  5. 17𝑛+1

Answer

We can solve this problem by substituting successive integer values of 𝑛 into the given general terms, beginning with 𝑛=1.

In option A, we substitute into 18𝑛, beginning with 𝑛=1, which gives the first term as 18×1=18.

Substituting 𝑛=2 to calculate the second term, we have 18×2=36.

However, the second term in the given sequence is not 36. Therefore, it is not described by the general term 18𝑛.

Next, we substitute 𝑛=1 into the general term in option B, 18𝑛, to give the first term as 18×1=18×1=18.

The second term, when 𝑛=2, is calculated by 18×2=18×4=72.

When 𝑛=3, we have 18×3=18×9=162.

And, the fourth term, when 𝑛=4, is 18×4=18×16=288.

As these terms match the first 4 terms of the given sequence, the sequence (18,72,162,288,) has the general term 18𝑛.

We can note that while the remaining three general terms in the options do give a correct first term of 18 in each case, the second and subsequent terms would not be equal to 72, 162, and 288. So, option B is the correct answer.

In the next example, we will find the general term of a more complex sequence involving powers of 3.

Example 5: Finding the General Term of a Sequence

The 𝑛th term of the sequence 3,92,9, is .

  1. 31
  2. 3
  3. 3𝑛
  4. 3𝑛

Answer

In order to find the 𝑛th term of the sequence, we can apply some logic by inspecting the sequence. The values 3, 9 (from the fraction 92), and 9 are powers of 3. Interestingly, we might notice that the second term has a denominator of 2. We may then consider if the terms of the sequence are all, in fact, fractions, for which some of the terms have been simplified to give a whole number.

We can consider if each term of the sequence has a numerator that is a power of 3 and a denominator that is the index.

Index, 𝑛123
33927
3𝑛3192273

We can simplify the values 31,92,273 to give 3,92,9.

These match the terms of the sequence that we were given; hence, we can give the answer that the 𝑛th term of the sequence is that given in option C, 3𝑛.

Note that we could have found the answer using the possible answer options. In this case, substituting the values 𝑛=1, 2, and 3, into each 𝑛th term would have given the result that the only 𝑛th term that produces the given sequence values 3,92,9 is 3𝑛.

In the next example, we will see a type of alternating sequence, which is defined below.

Definition: Alternating Sequences

An alternating sequence is one where the terms of the sequence alternate between positive and negative.

Example 6: Finding the General Term of an Alternating Sequence

The general term of the sequence 3,6,9,12,15, is 𝑇=.

  1. (1)×3𝑛
  2. (1)×3𝑛
  3. 3𝑛
  4. 3𝑛

Answer

We note that the sign of each term changes between successive terms, giving us alternating positive and negative values. This type of sequence is defined as an alternating sequence.

If we considered the absolute values of the terms, they would form a sequence 3,6,9,12,15. If the index 𝑛1, then this sequence of absolute values would be defined by the general term 𝑇=3𝑛.

One way to find the general term of a sequence that includes 3𝑛, but which alternates between positive and negative terms, is to multiply 3𝑛 by either (1) or (1). The general terms given in options A and B represent two alternatives.

In option A, for an index 𝑛1, we can substitute into the general term 𝑇=(1)×3𝑛, beginning with 𝑛=1. This gives 𝑇=(1)×3(1)=(1)×3=3.

We could continue to check further values of 𝑛, but we notice that the first term generated by the general term 𝑇=(1)×3𝑛 is 3, and not the value 3 from the given sequence. Therefore, this general term would not produce the given terms of the sequence, and we can eliminate answer option A.

Checking the 𝑛th term in answer option B, 𝑇=(1)×3𝑛, beginning with 𝑛=1, we have 𝑇=(1)×3(1)=(1)×3=1×3=3.

For the second term, we substitute 𝑛=2, giving 𝑇=(1)×3(2)=(1)×6=(1)×6=6.

Next, substituting 𝑛=3, we have 𝑇=𝑇(1)×3(3)=(1)×9=1×9=9.

Finally, substituting 𝑛=4 and 𝑛=5 gives 𝑇=(1)×3(4)=(1)×12=(1)×12=12 and 𝑇=(1)×3(5)=(1)×15=1×15=15.

These first 5 terms, 3,6,9,12, and 15, match the given terms of the sequence. Therefore, we can give the answer that the general term of the sequence is 𝑇=(1)×3𝑛.

In the final example, we will identify the recursive formula for the given terms of a sequence.

Example 7: Finding a Recursive Formula of a Sequence

Consider the sequence 4,10,22,46,. Which of the following recursive formulas can be used to calculate successive terms of the sequence for an index 𝑛1?

  1. 𝑇=4;𝑇=2𝑇
  2. 𝑇=4;𝑇=2𝑇+2
  3. 𝑇=4;𝑇=52𝑇
  4. 𝑇=4;𝑇=𝑇+6

Answer

We can recall that a recursive formula is a formula in which the terms of a sequence are defined using one or more previous terms. Here, in order to find a term with index 𝑛+1, we need to know the previous term with index 𝑛, 𝑇.

We can use each given recursive formula to generate a sequence and establish which matches the given sequence.

Beginning with the recursive formula in option A, we are given the first term, 𝑇=4. Therefore, to find the second term, we substitute 𝑛=1 and 𝑇=4 into the formula 𝑇=2𝑇, giving 𝑇=2𝑇=2(4)=8.

The sequence generated by the recursive formula in option A would begin 4,8, and as such does not match the given sequence. Therefore, we can eliminate this formula.

To check the recursive formula in option B, we have a first term 𝑇=4. To find the second term, we substitute 𝑛=1 and 𝑇=4 into the formula 𝑇=2𝑇+2, which gives 𝑇=2𝑇+2=2(4)+2=8+2=10.

This is the same as the second term in the given sequence, so we can continue to check the third term. Substituting 𝑛=2 and 𝑇=10, we have 𝑇=2𝑇+2=2(10)+2=20+2=22.

To check the fourth term, we substitute this value, 𝑇=22, and 𝑛=3, giving 𝑇=2𝑇+2=2(22)+2=44+2=46.

As these four terms match the given terms, the recursive formula for the sequence 4,10,22,46, is 𝑇=4;𝑇=2𝑇+2.

As an aside, if we wanted to explain this recursive formula in words, we would describe each term as double the previous term, plus 2. Considering the recursive formula in option C, we could describe each term as multiplying the previous term by the fraction 52. This would not generate the given terms of the sequence. Furthermore, the recursive formula in option D would describe a sequence where each term is found by adding 6 to the previous term. This would also not generate the given terms. Therefore, the formula given in option B is the correct one.

We now summarize the key points.

Key Points

  • To find the terms of a sequence given a general term, we substitute values of 𝑛1 into the formula for the general term.
  • A recursive formula is a formula in which the terms of a sequence are defined using one or more previous terms.
  • To find a specific term in a sequence using a recursive formula, we may need to apply the formula several times in order to find the values of preceding terms.
  • A sequence of real numbers (𝑇) is said to be increasing if 𝑇>𝑇 for all 𝑛.
  • A sequence of real numbers (𝑇) is said to be decreasing if 𝑇<𝑇 for all 𝑛.
  • A sequence of real numbers (𝑇) is said to be constant if 𝑇=𝑇 for all 𝑛.
  • An alternating sequence has terms that alternate between positive and negative values.

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