### 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 π’ sub one, π’ sub two, π’ sub three, and so on, all the way up
to π’ sub π, then we can write the sum of these numbers as π’ sub one plus π’ sub
two plus π’ sub three, and so on, all the way up to π’ sub π.

If we let π’ sub π represent the
general term, then this can be rewritten as the sum of π’ sub π from π equals one
to π equals π. The symbol here is the Greek
capital letter π΄, 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 π’ sub π where π takes values from one to π. It is important to note that our
lower limit does not have to be equal to one. We can also write the sum of π’ sub
π where π takes values from π to π. π is the lower limit and π 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 πth power minus 52, where π takes values from one to four.

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

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

The values four and nine are the
lower and upper limits of π, respectively. The Greek letter π΄ means the sum
of. In this question, we need to find
the sum of six terms when π 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
minus one where π takes values from four to nine. We now need to substitute each of
the values of π into our expression. Four minus one is equal to
three. So we have two cubed. When π 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 minus one where π 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 within the expression weβre trying to sum. However, in this case, we just have
a constant, negative 25.

Evaluate the sum of negative 25
where π 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 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 πth 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 π’ sub π and the number of terms in the series π. 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 where π is a variable going from one to
20.

54 multiplied by 12 is equal to
648. Therefore, 54 multiplied by 12π is
648π. Using sigma notation, we can
therefore express the series as the sum of 648π where π 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 π’ sub π where π
takes values from one to π. We need to work out the expression
for π’ sub π, which is the general term of the series. At first glance, there doesnβt
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 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 squared, where π takes values from one to
eight. We could check this answer by
substituting in each of our values of π 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 π’ sub π where π takes values from
one to π. In this question, we need to find
an expression for this general term π’ sub π as well as the value of π. Letβs 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 where π 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, 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 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 plus
one multiplied by π plus 495, where π takes values from one to 36.

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

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