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Lesson Video: Evaluating Series Algebraically Mathematics

In this video, we will learn how to evaluate linear and quadratic series by applying algebraic methods and formulas.

18:00

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 π‘Žπ‘› plus 𝑏. The general term of a quadratic sequence is π‘Žπ‘› squared plus 𝑏𝑛 plus 𝑐, where π‘Ž 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‘Ž sub π‘Ÿ, which is a function of π‘Ÿ, from π‘Ÿ equals one to some value 𝑛 using the sigma notation denoted as shown. The sum of π‘Ž sub π‘Ÿ from π‘Ÿ equals one to 𝑛 is equal to π‘Ž sub one plus π‘Ž sub two plus π‘Ž sub three, and so on, all the way up to π‘Ž sub 𝑛. This series starts at π‘Ÿ equals one. However, this is not a necessity. For example, we could start at π‘Ÿ equals π‘š, where π‘š is less than 𝑛, as shown.

The sum of π‘Ž sub π‘Ÿ from π‘Ÿ equals π‘š to π‘Ÿ equals 𝑛 can also be written as the sum of π‘Ž sub π‘Ÿ from π‘Ÿ equals one to 𝑛 minus the sum of π‘Ž sub π‘Ÿ from π‘Ÿ equals one to π‘š minus one. This is because the first series contains all the terms from π‘Ž sub one up to π‘Ž sub 𝑛. Subtracting the terms from π‘Ÿ equals one to π‘Ÿ equals π‘š minus one leaves us with π‘Ž sub π‘š plus π‘Ž sub π‘š plus one, and so on, all the way up to π‘Ž sub 𝑛. 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œ† multiplied by π‘Ž sub π‘Ÿ from π‘Ÿ equals one to 𝑛 is equal to πœ† multiplied by the sum of π‘Ž sub π‘Ÿ between π‘Ÿ equals one and 𝑛. This property holds if πœ† is a real constant. Next, we have that the sum of π‘Ž sub π‘Ÿ plus 𝑏 sub π‘Ÿ from π‘Ÿ equals one to 𝑛 is equal to the sum of π‘Ž sub π‘Ÿ from π‘Ÿ equals one to 𝑛 plus the sum of 𝑏 sub π‘Ÿ from π‘Ÿ equals one to 𝑛. These two properties can be combined. The sum of πœ† one π‘Ž sub π‘Ÿ plus πœ† two 𝑏 sub π‘Ÿ from π‘Ÿ equals one to 𝑛 is equal to πœ† one multiplied by the sum of π‘Ž sub π‘Ÿ from π‘Ÿ equals one to 𝑛 plus πœ† two multiplied by the sum of 𝑏 sub π‘Ÿ from π‘Ÿ equals one to 𝑛. 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‘š, which is greater than one and less than 𝑛. The sum of π‘Ž sub π‘Ÿ between π‘Ÿ equals one and 𝑛 is equal to the sum of π‘Ž sub π‘Ÿ between π‘Ÿ equals one and π‘š plus the sum of π‘Ž sub π‘Ÿ from π‘Ÿ equals π‘š plus one to 𝑛. This property holds as the first series on the right-hand side is equal to π‘Ž sub one plus π‘Ž sub two, and so on, up to π‘Ž sub π‘š. The second series is the sum of the remaining terms all the way up to π‘Ž sub 𝑛. 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‘Ÿ plus one from π‘Ÿ equals one to 12 plus the sum of four π‘Ÿ plus one from π‘Ÿ equals 13 to 25 is equal to what. Is it (A) the sum of eight π‘Ÿ plus two from π‘Ÿ equals one to 25, (B) the sum of four π‘Ÿ plus one squared from π‘Ÿ equals one to 25, (C) the sum of four π‘Ÿ plus two from π‘Ÿ equals one to 25, or (D) the sum of four π‘Ÿ plus one from π‘Ÿ equals one to 25?

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

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

Next, we will consider how we can find the sum of some linear value π‘Ÿ from π‘Ÿ equals one to 𝑛. This is equal to the sum of integers from one to 𝑛. 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‘Ÿ from π‘Ÿ equals one to 𝑛 is equal to 𝑛 plus one plus 𝑛 plus one, and so on, where there are 𝑛 lots of 𝑛 plus one. The right-hand side is therefore equal to 𝑛 multiplied by 𝑛 plus one. Dividing through by two, the sum of π‘Ÿ from π‘Ÿ equals one to 𝑛 is equal to 𝑛 multiplied by 𝑛 plus one all divided by two. We will now use these two formulae to find the sum of a series in the form π›Όπ‘Ÿ plus 𝛽.

Find the sum of π‘Ÿ minus eight from π‘Ÿ equals one to 𝑛 given the sum of π‘Ÿ from π‘Ÿ equals one to 𝑛 is equal to 𝑛 multiplied by 𝑛 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‘Ÿ from π‘Ÿ equals one to 𝑛 minus eight multiplied by the sum of one from π‘Ÿ equals one to 𝑛. We are given in the question that the first term on the right-hand side is equal to 𝑛 multiplied by 𝑛 plus one divided by two. Recalling that the sum of one from π‘Ÿ equals one to 𝑛 is equal to 𝑛, the second term on the right-hand side is equal to eight 𝑛. We have 𝑛 multiplied by 𝑛 plus one over two minus eight 𝑛.

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

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

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

We will begin by distributing our parentheses so that the linear expression is equal to nine π‘Ÿ minus 333. We need to calculate the sum of this from π‘Ÿ equals eight to π‘Ÿ 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‘š is equal to eight and 𝑛 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‘Ÿ from π‘Ÿ equals one to 12 minus 333 multiplied by the sum of one from one to 12 minus nine multiplied by the sum of π‘Ÿ from π‘Ÿ equals one to seven plus 333 multiplied by the sum of one from π‘Ÿ equals one to seven.

We are given an expression for the sum of π‘Ÿ from π‘Ÿ equals one to 𝑛. And in this question, 𝑛 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‘Ÿ equals one to 𝑛 is equal to 𝑛. 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‘Ÿ minus 37 from π‘Ÿ 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, let’s consider what happens when we have a quadratic expression. As for the sum of a constant and the linear term π‘Ÿ, there is a formula we can use if we need to calculate the sum of π‘Ÿ squared from π‘Ÿ equals one to 𝑛. 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‘Ÿ squared from π‘Ÿ equals one to 𝑛 is equal to 𝑛 multiplied by 𝑛 plus one multiplied by two 𝑛 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‘Ÿ from π‘Ÿ equals one to 𝑛 is equal to 𝑛 multiplied by 𝑛 plus one divided by two and the sum of π‘Ÿ squared from π‘Ÿ equals one to 𝑛 is equal to 𝑛 multiplied by 𝑛 plus one multiplied by two 𝑛 plus one all divided by six, use the properties of summation notation sigma to find the sum of seven π‘Ÿ squared minus seven π‘Ÿ minus 21 from π‘Ÿ 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‘Ÿ squared from π‘Ÿ equals one to four minus seven multiplied by the sum of π‘Ÿ from π‘Ÿ equals one to four minus 21 multiplied by the sum of one from π‘Ÿ equals one to four. We are given expressions in terms of 𝑛 for the sum of π‘Ÿ and the sum of π‘Ÿ squared. We also recall that the sum of one from π‘Ÿ equals one to 𝑛 is equal to 𝑛.

In this question, the value of 𝑛 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‘Ÿ squared minus seven π‘Ÿ minus 21 from π‘Ÿ equals one to four is 56.

Alternatively, we could’ve substituted the integers from one to four into our expression and then found the sum of these four values. When π‘Ÿ equals one, seven π‘Ÿ squared minus seven π‘Ÿ minus 21 is equal to negative 21. When π‘Ÿ equals two, the expression equals negative seven. When π‘Ÿ equals three, we get an answer of 21. And when π‘Ÿ 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, we’ve evaluated series written in the form the sum of π›Όπ‘Ÿ squared plus π›½π‘Ÿ plus 𝛾 from π‘Ÿ equals π‘š to 𝑛, noting that when 𝛼 equals zero, we have a linear series and when 𝛼 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‘Ÿ equals one to 𝑛 is equal to 𝑛. The sum of π‘Ÿ from π‘Ÿ equals one to 𝑛 is equal to 𝑛 multiplied by 𝑛 plus one divided by two. And the sum of π‘Ÿ squared from π‘Ÿ equals one to 𝑛 is equal to 𝑛 multiplied by 𝑛 plus one multiplied by two 𝑛 plus one all divided by six.

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