### 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 πth term and written π sub π, is an algebraic expression
that relates the term to its position number in the sequence. Letβs consider the general term π
sub π equal to three π 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 π is equal to one, two, and three. When π 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 sub one is equal
to seven, π sub two is equal to 10, and π sub three is equal to 13. The sequence with general term π
sub π equals three π plus four is seven, 10, 13, and so on.

Letβs now consider how we could
calculate the eighth term in this sequence. Substituting π equals eight into
our expression, we have three multiplied by eight plus four. This is equal to 28. π 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 π, we substitute the term number in place of π to find a
specific term in the sequence. For example, to find the 20th term,
we substitute π 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 πth term is given by π sub π equals π squared minus 14, where π
is greater than or equal to one.

In order to calculate the first
five terms of our sequence, we need to substitute π equals one, two, three, four,
and five into our expression for π sub π. The first term is when π equals
one. We denote this as π 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 π 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 π equals three, π sub three
is equal to negative five. When π equals four, four squared
minus 14 equals two. And finally, the fifth term π sub
five is equal to five squared minus 14, which equals 11. The first five terms of a sequence
whose πth term is given by π sub π equals π 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 sub π is equal to π cubed minus 14.

The seventh term of a sequence with
a general term π sub π will be denoted π sub seven. We can calculate this by
substituting π equals seven into our expression for the general term. π 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
π. For example, to calculate the 20th
term, we would substitute π 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. Letβs consider the general term
with recursive formula π sub π is equal to two multiplied by π sub π minus one
plus five, where π is greater than or equal to two and π sub one is equal to
four.

We see that the expression here
contains the term π sub π minus one. This is the term immediately prior
π sub π. 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 sub one is
equal to four. Therefore, when π is equal to one,
π sub π 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 sub π is equal to two multiplied by π sub π 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 sub π minus one, we substitute the preceding term in
place of π sub π minus one to find any term. If, however, the general term
contains an expression for π sub π plus one in π sub π, we substitute the
preceding term in place of π sub π to find the value of π sub π 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 sub π, given π sub π plus one is equal to negative one to the πth
power divided by nine multiplied by π sub π, where π is greater than or equal to
one and π sub one equals negative 11.

In this question, we are given the
value of the first term of our sequence π 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 sub two, π
sub three, π sub four, and π sub five.

The formula we are given is an
example of a recursive formula, where π sub π plus one is given in terms of π sub
π. We can therefore calculate any term
of the sequence by substituting in the previous term. The second term π sub two is
therefore equal to negative one to the power of one divided by nine multiplied by π
sub one. We know that π 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 sub three is equal to negative one squared over nine multiplied by π sub
two. We know that π 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 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 sub three, the third term in our
sequence, is 11.

Next, we need to calculate the
fourth term π sub four. This is equal to negative one cubed
divided by nine multiplied by π sub three. And we have just calculated that π
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 sub five is
equal to negative one to the fourth power divided by nine multiplied by π sub
four. We can substitute negative one over
99 for π sub four. And this in turn simplifies to one
over negative one eleventh. Using a similar method that we used
to calculate π sub three, we have π 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 sub one is equal to π sub five,
which is equal to π 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
π minus one? (B) 52 plus 32 multiplied by π
minus one. (C) 52 plus 32 multiplied by π
plus one. (D) 84 plus 32 multiplied by π
minus one. Or (E) 84 plus 30 multiplied by π
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 π, 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. Letβs begin with π 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 π 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 π 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
π equals three and π equals four. When π is equal to three, the
expression 52 plus 32 multiplied by π minus one gives us 116. And when π 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 π, then we substitute the term number in place of π. However, if the general term
contains an expression in π sub π minus one, we substitute the preceding term in
place of π sub π minus one.