Lesson Explainer: Evaluating Series Algebraically | Nagwa Lesson Explainer: Evaluating Series Algebraically | Nagwa

Lesson Explainer: Evaluating Series Algebraically Mathematics • Second Year of Secondary School

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

We will begin by recalling what we mean by a series as the sum of terms in a sequence {𝑎} and how we use sigma notation to represent this.

Definition: Series Using Sigma Notation

We can take the sum of a term 𝑎, which is a function of 𝑟, from 𝑟=1 to some value 𝑛 using the sigma notation denoted by : 𝑎=𝑎+𝑎+𝑎++𝑎+𝑎.

The series can also start from a different value, from 𝑟=𝑚 for 𝑚<𝑛 to some value 𝑛, which can be written as 𝑎=𝑎+𝑎++𝑎+𝑎=𝑎𝑎, which is the difference of two series with a starting index 𝑟=1.

In order to evaluate these series algebraically, we will make use of some properties using the notation. These are intuitive properties which are reminiscent of how we work with algebraic expressions, but it is useful to familiarize ourselves with the more formal notation.

Definition: Properties of Series

The sums of the series written in terms of the sigma notation satisfy the following properties:

  • If we have a constant 𝜆 appearing inside the sum, we can take it out of the sum as follows: 𝜆𝑎=𝜆𝑎.
  • We can also split up the sum for different terms in the summand as follows: (𝑎+𝑏)=𝑎+𝑏.
  • We can also split up the summation for some value in between the start and end, 1<𝑚<𝑛. If we start at 𝑟=1 and sum up to 𝑛, then for 𝑚<𝑛 we have 𝑎=𝑎+𝑎.

The first two properties show that the summation is linear and we can combine them as (𝜆𝑎+𝜆𝑏)=𝜆𝑎+𝜆𝑏.

For the last property, we can also see this by using the property for a starting index greater than 1, as follows: 𝑎+𝑎=𝑎+𝑎𝑎=𝑎.

In this explainer, we will look at linear or quadratic series, that is, where 𝑎 is a linear or quadratic function of 𝑟.

Let’s consider an example where we use the last property to rewrite a summation.

Example 1: Simplifying a Finite Series Using the Properties of Summation

Fill in the blank: (4𝑟+1)+(4𝑟+1)=.

  1. (8𝑟+2)
  2. (4𝑟+1)
  3. (4𝑟+2)
  4. (4𝑟+1)

Answer

In this example, we will simplify a series algebraically.

We will make use of the property for starting indices greater than 1: 𝑎=𝑎𝑎.

For the given summation, we have (4𝑟+1)+(4𝑟+1)=(4𝑟+1)+(4𝑟+1)(4𝑟+1)=(4𝑟+1)+(4𝑟+1)(4𝑟+1)=(4𝑟+1).

This makes sense, since the first term is the sum from 𝑟=1 to 12, while the second sum is from 𝑟=13 to 25; thus, the total sum will be from 𝑟=1 to 25. If we apply the property where we can split the series for some 𝑚<𝑛 as 𝑎=𝑎+𝑎, we obtain the same result upon substituting 𝑎=4𝑟+1 and 𝑚=12, which gives the given series on the right-hand side and the result on the left-hand side.

This is option D.

Now let’s calculate the summation of a series with a constant inside the summand, which we can take outside the sum: 𝑆=𝛼=𝛼1.

Note that 1=1+1++1+1=𝑛.times

Thus for 𝑆, the summation for a constant series, we have 𝑆=𝛼1=𝛼𝑛.

Now, let’s look at an example where we evaluate an arithmetic series with a constant summand.

Example 2: Evaluating Arithmetic Series

Evaluate 5.

Answer

In this example, we want to evaluate an arthmetic series with a constant summand.

We will make use of the following summation: 𝛼=𝛼𝑛.

Thus, for the given sum, we have 𝛼=5 and 𝑛=9 and hence 5=5×9=45.

Now, let’s consider a linear summation of the form 𝑆=(𝛼𝑟+𝛽).

By the property for adding series and the constant multiple property, we can rewrite this as 𝑆=𝛼𝑟+𝛽1.

In order to evaluate this sum, we first need a formula for the series 𝑟=1+2++𝑛.

We can write this series out explicitly in two different ways: by starting from 1 and summing all the numbers up to 𝑛 and by starting from 𝑛 and summing all numbers down to 1: 𝑟=1+2+3++(𝑛1)+𝑛,𝑟=𝑛+(𝑛1)+(𝑛2)2+1.

Adding these together gives 2𝑟=(𝑛+1)+(𝑛1+2)+(𝑛2+3)(𝑛1+2)+(𝑛+1)=(𝑛+1)+(𝑛+1)+(𝑛+1)+(𝑛+1)=𝑛(𝑛+1).times

Thus, we have 𝑟=𝑛(𝑛+1)2.

We can also derive this from the binomial expansion of (𝑟1)=𝑟2𝑟+1, which upon rearranging gives 𝑟(𝑟1)=2𝑟1.

First note that the summation of the left-hand side from 𝑟=1 to 𝑛 gives 𝑟(𝑟1)=10+21+32++𝑛(𝑛1)=11+22+33+(𝑛1)(𝑛1)+𝑛=𝑛, while the right-hand side gives (2𝑟1)=2𝑟1=2𝑟𝑛.

Thus, we obtain 2𝑟𝑛=𝑛2𝑟=𝑛+𝑛=𝑛(𝑛+1).

This gives the same formula as before: 𝑟=𝑛(𝑛+1)2.

Using this result, we can now evaluate the series of the linear expression defined by 𝑆: 𝑆=𝛼𝑟+𝛽1=𝛼𝑛(𝑛+1)2+𝛽𝑛=𝛼𝑛(𝑛+1)+2𝛽𝑛2=𝑛(𝛼(𝑛+1)+2𝛽)2.

In the next example, we will look at evaluating a sum algebraically using the properties of summation for a series with a linear summand.

Example 3: Evaluating the Sum of an Finite Series Using the Properties of Summation

Find (𝑟8) given 𝑟=𝑛(𝑛+1)2.

Answer

In this example, we will evaluate a linear series algebraically using properties of summation.

We will make use of the summation linearity property (𝜆𝑎+𝜆𝑏)=𝜆𝑎+𝜆𝑏, and the summations 1=𝑛,𝑟=𝑛(𝑛+1)2.

The given summation can be written as (𝑟8)=𝑟8=𝑟81=𝑛(𝑛+1)28𝑛=𝑛+𝑛16𝑛2=𝑛15𝑛2.

Now, let’s look at an example where we determine the value of an unknown appearing in a linear series with a given value for the sum.

Example 4: Finding an Unknown when Given the Value of an Arithmetic Series

If (3+𝑘𝑟)=150, find 𝑘.

  1. 19093
  2. 193
  3. 1190
  4. 93190
  5. 3167

Answer

In this example, we want to find the value of an unknown appearing inside an arithmetic sum.

We will make use of the summation linearity property (𝜆𝑎+𝜆𝑏)=𝜆𝑎+𝜆𝑏, and the summations 1=𝑛,𝑟=𝑛(𝑛+1)2.

Let’s first evaluate the right-hand side of the given sum: (3+𝑘𝑟)=3+𝑘𝑟=31+𝑘𝑟=3×19+𝑘×19(19+1)2=57+190𝑘.

Using this, we can determine the value of 𝑘 as 57+190𝑘=150190𝑘=93𝑘=93190.

Now, let’s evaluate a linear series with a starting index greater than 1.

Example 5: Evaluating Arithmetic Series with a Starting Index Greater Than 1

Find 9(𝑟37) using the properties of summation and given 𝑟=𝑛(𝑛+1)2.

Answer

In this example, we will evaluate a linear arithmetic series with a starting index greater than 1 using the properties of summation.

We will make use of the property for starting indices greater than 1 and the summation linearity property 𝑎=𝑎𝑎,(𝜆𝑎+𝜆𝑏)=𝜆𝑎+𝜆𝑏, and the summations 1=𝑛,𝑟=𝑛(𝑛+1)2.

Using the properties, we can rewrite the given sum as 9(𝑟37)=(9𝑟333)=(9𝑟333)(9𝑟333)=9𝑟33319𝑟+3331=9×12(12+1)2333×129×7(7+1)2+333×7=9×78333×129×28+333×7=9×50333×5=1215.

Now, let’s consider a quadratic expression where we want to evaluate the series: 𝑆=𝛼𝑟+𝛽𝑟+𝛾.

By the property for adding series and the constant multiple property, we can rewrite this as 𝑆=𝛼𝑟+𝛽𝑟+𝛾1.

In order to evaluate this sum, we first need a formula for the series 𝑟=1+2++𝑛.

Similar to the linear term, we will consider the binomial expansion of (𝑟1)=𝑟3𝑟+3𝑟1, which upon rearranging can be written as 𝑟(𝑟1)=3𝑟3𝑟+1.

The summation of the left-hand side from 𝑟=1 to 𝑛 gives 𝑟(𝑟1)=10+21+32++𝑛(𝑛1)=11+22+33+(𝑛1)(𝑛1)+𝑛=𝑛, and from the right-hand side we have 3𝑟3𝑟+1=3𝑟3𝑟+1=3𝑟3𝑛(𝑛+1)2+𝑛.

Thus, we have 3𝑟3𝑛(𝑛+1)2+𝑛=𝑛,3𝑟=𝑛+3𝑛(𝑛+1)2𝑛=𝑛(𝑛+1)(2𝑛+1)2, and upon rearranging we obtain the formula 𝑟=𝑛(𝑛+1)(2𝑛+1)6.

Using this result, we can evaluate the series 𝑆 as follows: 𝑆=𝛼𝑟+𝛽𝑟+𝛾1=𝛼𝑛(𝑛+1)(2𝑛+1)6+𝛽𝑛(𝑛+1)2+𝛾𝑛=𝑛(𝛼(𝑛+1)(2𝑛+1)+3𝛽(𝑛+1)+6𝛾)6.

Let’s look at a few examples of how to evaluate a quadratic series. In the next example, we will evaluate a summand that contains a square and constant term using the properties of summation.

Example 6: Evaluating the Sum of a Finite Quadratic Series Using the Properties of Summation

Given 𝑟=𝑛(𝑛+1)(2𝑛+1)6, use the properties of summation to find 5𝑟67.

Answer

In this example, we will evaluate a quadratic series containing a square and constant term, using the properties of summation.

We will make use of the summation linearity property (𝜆𝑎+𝜆𝑏)=𝜆𝑎+𝜆𝑏, and the summations 1=𝑛,𝑟=𝑛(𝑛+1)(2𝑛+1)6.

Using these properties and summations, we can evaluate the given series: 5𝑟67=5𝑟671=5×6(6+1)(12+1)667×6=5×9167×6=53.

Now, let’s look at an example where we will evaluate a summand that contains a quadratic function of 𝑟.

Example 7: Evaluating the Sum of a Finite Quadratic Series Using the Properties of Summation

Given that 𝑟=𝑛(𝑛+1)2 and 𝑟=𝑛(𝑛+1)(2𝑛+1)6, use the properties of the summation notation to find 7𝑟7𝑟21.

Answer

In this example, we will evaluate a quadratic series using the properties of summation.

We will make use of the summation linearity property (𝜆𝑎+𝜆𝑏)=𝜆𝑎+𝜆𝑏, and the summations 1=𝑛,𝑟=𝑛(𝑛+1)2,𝑟=𝑛(𝑛+1)(2𝑛+1)6.

Using these properties and summations, we can evaluate the given series: 7𝑟7𝑟21=7𝑟7𝑟211=7×4(4+1)(8+1)67×4(4+1)221×4=7×307×1021×4=56.

Finally, let’s evaluate a quadratic series with a starting index greater than 1.

Example 8: Evaluating the Sum of a Finite Quadratic Series with a Starting Index Greater Than 1 Using the Properties of Summation

Given that 𝑟=𝑛(𝑛+1)2 and 𝑟=𝑛(𝑛+1)(2𝑛+1)6, use the properties of the summation notation to find 5𝑟22𝑟.

Answer

We will make use of the property for starting indices greater than 1 and the summation linearity property 𝑎=𝑎𝑎,(𝜆𝑎+𝜆𝑏)=𝜆𝑎+𝜆𝑏, and the summations 𝑟=𝑛(𝑛+1)2,𝑟=𝑛(𝑛+1)(2𝑛+1)6.

Using these properties and summations, we can evaluate the given series: 5𝑟22𝑟=5𝑟22𝑟5𝑟22𝑟=5𝑟22𝑟5𝑟+22𝑟=5×8(8+1)(16+1)622×8(8+1)25×4(4+1)(8+1)6+22×4(4+1)2=5×20422×365×30+22×10=298.

Key Points

  • We have evaluated different series of the form 𝑆=𝛼𝑟+𝛽𝑟+𝛾, where 𝛼=0 for a linear series, while 𝛼0 for a quadratic series.
  • In order to evaluate these, we make use of the properties 𝑎=𝑎𝑎,(𝜆𝑎+𝜆𝑏)=𝜆𝑎+𝜆𝑏. The first allows us to evaluate a series with a starting index greater than one, by writing it as a difference of two series of a starting index equal to one. The second is a linearity property, which allows us to split the summation for different terms and take out the constant.
  • We also make use of the following series: 1=𝑛,𝑟=𝑛(𝑛+1)2,𝑟=𝑛(𝑛+1)(2𝑛+1)6.
  • When the starting index is equal to 1, that is, 𝑚=1, we have the following result for the general quadratic series: 𝑆=𝑛(𝛼(𝑛+1)(2𝑛+1)+3𝛽(𝑛+1)+6𝛾)6.

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