### Video Transcript

In this video, we will learn how to
evaluate linear and quadratic series by applying algebraic methods and formulae. We recall that a linear sequence
has a common difference and general term ππ plus π. The general term of a quadratic
sequence is ππ squared plus ππ plus π, where π is nonzero.

We will begin by recalling what we
mean by a series as the sum of the terms in a sequence and how we use sigma notation
to represent these. We can take the sum of a term π
sub π, which is a function of π, from π equals one to some value π using the
sigma notation denoted as shown. The sum of π sub π from π equals
one to π is equal to π sub one plus π sub two plus π sub three, and so on, all
the way up to π sub π. This series starts at π equals
one. However, this is not a
necessity. For example, we could start at π
equals π, where π is less than π, as shown.

The sum of π sub π from π equals
π to π equals π can also be written as the sum of π sub π from π equals one to
π minus the sum of π sub π from π equals one to π minus one. This is because the first series
contains all the terms from π sub one up to π sub π. Subtracting the terms from π
equals one to π equals π minus one leaves us with π sub π plus π sub π plus
one, and so on, all the way up to π sub π. This is known as the difference of
two series.

In order to evaluate these series
algebraically, we will now look at some properties using the sigma notation. These properties follow the same
method as when we are working with algebraic expressions.

Firstly, we have the sum of π
multiplied by π sub π from π equals one to π is equal to π multiplied by the
sum of π sub π between π equals one and π. This property holds if π is a real
constant. Next, we have that the sum of π
sub π plus π sub π from π equals one to π is equal to the sum of π sub π from
π equals one to π plus the sum of π sub π from π equals one to π. These two properties can be
combined. The sum of π one π sub π plus π
two π sub π from π equals one to π is equal to π one multiplied by the sum of
π sub π from π equals one to π plus π two multiplied by the sum of π sub π
from π equals one to π. This is known as the linear
property of summation.

Finally, we can also split up the
summation for some value between the start and end, in this case the value π, which
is greater than one and less than π. The sum of π sub π between π
equals one and π is equal to the sum of π sub π between π equals one and π plus
the sum of π sub π from π equals π plus one to π. This property holds as the first
series on the right-hand side is equal to π sub one plus π sub two, and so on, up
to π sub π. The second series is the sum of the
remaining terms all the way up to π sub π. Combining these gives us the sum on
the left-hand side. In our first example, we will use
this property to rewrite a summation.

The sum of four π plus one from π
equals one to 12 plus the sum of four π plus one from π equals 13 to 25 is equal
to what. Is it (A) the sum of eight π plus
two from π equals one to 25, (B) the sum of four π plus one squared from π equals
one to 25, (C) the sum of four π plus two from π equals one to 25, or (D) the sum
of four π plus one from π equals one to 25?

In order to answer this question,
we will use the property of series that states that the sum of π sub π from π
equals one to π plus the sum of π sub π from π equals π plus one to π is equal
to the sum of π sub π from π equals one to π. This holds when π is less than
π. From the expression given, we see
that the value of π is 12. π plus one is therefore equal to
13, which is the start value of our second summation. We can also see that π is equal to
25 and π sub π is equal to four π plus one. The expression on the left-hand
side of our equation is therefore equal to the sum of four π plus one from π
equals one to 25. This means that the correct answer
from the four options is option (D).

Letβs now consider how we can
calculate the summation of a series with a constant. Letβs assume we need to calculate
the sum of some constant πΌ from π equals one to π. We already know that we can take a
constant outside of the summation. So this is equal to πΌ multiplied
by the sum of one from π equals one to π. The summation part is equal to one
plus one plus one, and so on, where there are π ones. This is therefore equal to π. The sum of any constant πΌ from π
equals one to π is equal to πΌ multiplied by π.

Next, we will consider how we can
find the sum of some linear value π from π equals one to π. This is equal to the sum of
integers from one to π. As addition is commutative, we
could also write the terms in reverse order. We will then add these two
equations. This gives us two multiplied by the
sum of π from π equals one to π is equal to π plus one plus π plus one, and so
on, where there are π lots of π plus one. The right-hand side is therefore
equal to π multiplied by π plus one. Dividing through by two, the sum of
π from π equals one to π is equal to π multiplied by π plus one all divided by
two. We will now use these two formulae
to find the sum of a series in the form πΌπ plus π½.

Find the sum of π minus eight from
π equals one to π given the sum of π from π equals one to π is equal to π
multiplied by π plus one divided by two.

We begin this question by recalling
the linear property of summation. This means that we can rewrite the
summation given as the sum of π from π equals one to π minus eight multiplied by
the sum of one from π equals one to π. We are given in the question that
the first term on the right-hand side is equal to π multiplied by π plus one
divided by two. Recalling that the sum of one from
π equals one to π is equal to π, the second term on the right-hand side is equal
to eight π. We have π multiplied by π plus
one over two minus eight π.

Distributing the parentheses and
creating a common denominator, we have π squared plus π minus 16π all over
two. This in turn simplifies to π
squared minus 15π over two. This is the sum of π minus eight
from π equals one to π.

We will now look at an example
where the starting index is greater than one.

Find the sum of nine multiplied by
π minus 37 from π equals eight to 12 using the properties of summation and given
the sum of π from π equals one to π is π multiplied by π plus one divided by
two.

We will begin by distributing our
parentheses so that the linear expression is equal to nine π minus 333. We need to calculate the sum of
this from π equals eight to π equals 12 using the properties of summation.

When our starting index is greater
than one, we can use the difference of two series property. As our value of π is equal to
eight and π is equal to 12, we can rewrite the expression as shown. Next, we can use the linear
property of summation. Recalling that when subtracting a
negative number we get a positive answer, we are left with nine multiplied by the
sum of π from π equals one to 12 minus 333 multiplied by the sum of one from one
to 12 minus nine multiplied by the sum of π from π equals one to seven plus 333
multiplied by the sum of one from π equals one to seven.

We are given an expression for the
sum of π from π equals one to π. And in this question, π will be
equal to 12 and seven, respectively. This means that our first term is
equal to nine multiplied by 12 multiplied by 13 divided by two. And the third term is equal to nine
multiplied by seven multiplied by eight divided by two. These simplify to nine multiplied
by 78 and nine multiplied by 28.

We recall that the sum of one from
π equals one to π is equal to π. Our expression therefore becomes
nine multiplied by 78 minus 333 multiplied by 12 minus nine multiplied by 28 plus
333 multiplied by seven. This is equal to negative
1,215. This is the sum of nine multiplied
by π minus 37 from π equals eight to 12.

An alternative method here would be
to simply substitute the integers from eight to 12 into our expression and then find
the sum of these values.

Before looking at one final
example, letβs consider what happens when we have a quadratic expression. As for the sum of a constant and
the linear term π, there is a formula we can use if we need to calculate the sum of
π squared from π equals one to π. For the purpose of this video, we
will not consider the proof of this. However, it can be derived using
our knowledge of binomial expansions that the sum of π squared from π equals one
to π is equal to π multiplied by π plus one multiplied by two π plus one all
divided by six.

We will now look at one final
example where we use all three of these formulae to evaluate a quadratic series.

Given that the sum of π from π
equals one to π is equal to π multiplied by π plus one divided by two and the sum
of π squared from π equals one to π is equal to π multiplied by π plus one
multiplied by two π plus one all divided by six, use the properties of summation
notation sigma to find the sum of seven π squared minus seven π minus 21 from π
equals one to four.

We will begin by rewriting our
expression using the linear property of summation. This gives us seven multiplied by
the sum of π squared from π equals one to four minus seven multiplied by the sum
of π from π equals one to four minus 21 multiplied by the sum of one from π
equals one to four. We are given expressions in terms
of π for the sum of π and the sum of π squared. We also recall that the sum of one
from π equals one to π is equal to π.

In this question, the value of π
is four. The first term in our expression
becomes seven multiplied by four multiplied by five multiplied by nine all divided
by six. This can be simplified as shown,
giving us seven multiplied by 30. The second term is equal to seven
multiplied by four multiplied by five divided by two. This is simply seven multiplied by
10. As the third term becomes 21
multiplied by four, we are left with seven multiplied by 30 minus seven multiplied
by 10 minus 21 multiplied by four. This is equal to 56. The sum of seven π squared minus
seven π minus 21 from π equals one to four is 56.

Alternatively, we couldβve
substituted the integers from one to four into our expression and then found the sum
of these four values. When π equals one, seven π
squared minus seven π minus 21 is equal to negative 21. When π equals two, the expression
equals negative seven. When π equals three, we get an
answer of 21. And when π equals four, the
expression equals 63. As these four numbers sum to give
us 56, this confirms our answer is correct.

We will now summarize the key
points from this video. In this video, weβve evaluated
series written in the form the sum of πΌπ squared plus π½π plus πΎ from π equals
π to π, noting that when πΌ equals zero, we have a linear series and when πΌ is
not equal to zero, we have a quadratic series. We made use of two properties of
series: firstly, the difference property. Secondly, we used the linear
property of summation. We also used the fact that the sum
of one from π equals one to π is equal to π. The sum of π from π equals one to
π is equal to π multiplied by π plus one divided by two. And the sum of π squared from π
equals one to π is equal to π multiplied by π plus one multiplied by two π plus
one all divided by six.