An arithmetic series is the sum of the terms of an arithmetic sequence. The nth partial sum of an arithmetic sequence can be calculated using the first and last terms as follows: Sn=n(a1+an)2.
- 1 What is a partial sum arithmetic?
- 2 What is the partial sum of a sequence?
- 3 How do you find the nth partial sum of a sequence?
- 4 What is the formula of the sum of arithmetic sequence?
- 5 Introduction to Arithmetic Progressions
- 6 Finite Sequence: Definition & Examples – Video & Lesson Transcript
- 7 Examples
- 8 Finding Patterns
- 9 Finding Sums of Finite Arithmetic Series – Sequences and Series (Algebra 2)
- 10 Arithmetic Series
- 11 Partial Sums of an Arithmetic Sequence
- 12 Partial Sums
- 13 Sigma
- 14 More Powerful
- 15 Properties
- 16 Useful Shortcuts
- 17 Arithmetic Sequences and Sums
- 18 Arithmetic Sequence
- 19 Advanced Topic: Summing an Arithmetic Series
- 20 Footnote: Why Does the Formula Work?
What is a partial sum arithmetic?
The simple answer is that a partial sum is actually just the sum of part of a sequence. You can find a partial sum for both finite sequences and infinite sequences. A partial sum, on the other hand, is just the sum of part of a sequence.
What is the partial sum of a sequence?
A Partial Sum is the sum of part of the sequence. The sum of infinite terms is an Infinite Series. And Partial Sums are sometimes called “Finite Series”.
How do you find the nth partial sum of a sequence?
The nth partial sum of a geometric sequence can be calculated using the first term a1 and common ratio r as follows: Sn=a1(1−rn)1−r. The infinite sum of a geometric sequence can be calculated if the common ratio is a fraction between −1 and 1 (that is |r|<1) as follows: S∞=a11−r.
What is the formula of the sum of arithmetic sequence?
The sum of the arithmetic sequence can be derived using the general arithmetic sequence, an n = a1 1 + (n – 1)d.
Introduction to Arithmetic Progressions
Generally speaking, a progression is a sequence or series of numbers in which the numbers are organized in a certain order so that the relationship between the succeeding terms of a series or sequence remains constant. It is feasible to find the n thterm of a series by following a set of steps. There are three different types of progressions in mathematics:
- Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP) are all types of progression.
AP, also known as Arithmetic Sequence, is a sequence or series of integers in which the common difference between two subsequent numbers in the series is always the same. As an illustration: Series 1: 1, 3, 5, 7, 9, and 11. Every pair of successive integers in this sequence has a common difference of 2, which is always true. Series 2: numbers 28, 25, 22, 19, 16, 13,. There is no common difference between any two successive numbers in this series; instead, there is a strict -3 between any two consecutive numbers.
Terminology and Representation
- Common difference, d = a 2– a 1= a 3– a 2=. = a n– a n – 1
- A n= n thterm of Arithmetic Progression
- S n= Sum of first n elements in the series
- A n= n
General Form of an AP
Given that ais treated as a first term anddis treated as a common difference, the N thterm of the AP may be calculated using the following formula: As a result, using the above-mentioned procedure to compute the n terms of an AP, the general form of the AP is as follows: Example: The 35th term in the sequence 5, 11, 17, 23,. is to be found. Solution: When looking at the given series,a = 5, d = a 2– a 1= 11 – 5 = 6, and n = 35 are the values. As a result, we must use the following equations to figure out the 35th term: n= a + (n – 1)da n= 5 + (35 – 1) x 6a n= 5 + 34 x 6a n= 209 As a result, the number 209 represents the 35th term.
Sum of n Terms of Arithmetic Progression
The arithmetic progression sum is calculated using the formula S n= (n/2)
Derivation of the Formula
Allowing ‘l’ to signify the n thterm of the series and S n to represent the sun of the first n terms of the series a, (a+d), (a+2d),., a+(n-1)d S n = a 1 plus a 2 plus a 3 plus .a n-1 plus a n S n= a + (a + d) + (a + 2d) +. + (l – 2d) + (l – d) + l. + (l – 2d) + (l – d) + l. + (l – 2d) + (l – d) + l. (1) When we write the series in reverse order, we obtain S n= l + (l – d) + (l – 2d) +. + (a + 2d) + (a + d) + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a … (2) Adding equations (1) and (2) results in equation (2).
+ (a + l) + (a + l) + (a + l) +.
(3) As a result, the formula for calculating the sum of a series is S n= (n/2)(a + l), where an is the first term of the series, l is the last term of the series, and n is the number of terms in the series.
d S n= (n/2)(a + a + (n – 1)d)(a + a + (n – 1)d) S n= (n/2)(2a + (n – 1) x d)(n/2)(2a + (n – 1) x d) Observation: The successive terms in an Arithmetic Progression can alternatively be written as a-3d, a-2d, a-d, a, a+d, a+2d, a+3d, and so on.
Sample Problems on Arithmetic Progressions
Problem 1: Calculate the sum of the first 35 terms in the sequence 5,11,17,23, and so on. a = 5 in the given series, d = a 2–a in the provided series, and so on. The number 1 equals 11 – 5 = 6, and the number n equals 35. S n= (n/2)(2a + (n – 1) x d)(n/2)(2a + (n – 1) x d) S n= (35/2)(2 x 5 + (35 – 1) x 6)(35/2)(2 x 5 + (35 – 1) x 6) S n= (35/2)(10 + 34 x 6) n= (35/2)(10 + 34 x 6) n= (35/2)(10 + 34 x 6) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) A = 35214/2A = 3745S n= 35214/2A = 3745 Find the sum of a series where the first term of the series is 5 and the last term of the series is 209, and the number of terms in the series is 35, as shown in Problem 2.
S n= (35/2)(5 + 209) S n= (35/2)(5 + 209) S n= (35/2)(5 + 209) A = 35214/2A = 3745S n= 35214/2A = 3745 Problem 3: A amount of 21 rupees is divided among three brothers, with each of the three pieces of money being in the AP and the sum of their squares being the sum of their squares being 155.
Solution: Assume that the three components of money are (a-d), a, and (a+d), and that the total amount allocated is in AP.
155 divided by two equals 155 Taking the value of ‘a’ into consideration, we obtain 3(7) 2+ 2d.
2= 4d = 2 = 2 The three portions of the money that was dispersed are as follows:a + d = 7 + 2 = 9a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5 As a result, the most significant portion is Rupees 9 million.
Finite Sequence: Definition & Examples – Video & Lesson Transcript
When there is a finite series, the first term is followed by a second term, and so on until the last term. In a finite sequence, the letternoften reflects the total number of phrases in the sequence. In a finite series, the first term may be represented by a (1), the second term can be represented by a (2), etc. Parentheses are often used to separate numbers adjacent to thea, but parentheses will be used at other points in this course to distinguish them from subscripts. This terminology is illustrated in the following graphic.
Despite the fact that there are many other forms of finite sequences, we shall confine ourselves to the field of mathematics for the time being. The prime numbers smaller than 40, for example, are an example of a finite sequence, as seen in the table below: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, Another example is the range of natural numbers between zero and one hundred. Due to the fact that it would be tedious to write down all of the terms in this finite sequence, we will demonstrate it as follows: 1, 2, 3, 4, 5,., 100 are the numbers from 1 to 100.
Alternatively, there might be alternative sequences that begin in the same way and end in 100, but which do not contain the natural numbers less than or equal to 100.
Let’s take a look at some finite sequences to see if there are any patterns.
Examine a few finite sequences to see if any patterns can be discovered.
Finding Sums of Finite Arithmetic Series – Sequences and Series (Algebra 2)
The sum of all terms for a finitearithmetic sequencegiven bywherea1 is the first term,dis the common difference, andnis the number of terms may be determined using the following formula: wherea1 is the first term,dis the common difference, andnis the number of terms Consider the arithmetic sum as an example of how the total is computed in this manner. If you choose not to add all of the words at once, keep in mind that the first and last terms may be recast as2fives. This method may be used to rewrite the second and second-to-last terms as well.
There are 9fives in all, and the aggregate is 9 x 5 = 9.
expand more Due to the fact that the difference between each term is constant, this sequence from 1 through 1000 is arithmetic.
The first phrase, a1, is one and the last term, is one thousand thousand. The total number of terms is less than 1000. The sum of all positive integers up to and including 1000 is 500 500. }}}!}}}Test}}} arrow back} arrow forwardarrow leftarrow right
It is the sum of the terms of an arithmetic sequence that is known as an arithmetic series. A geometric series is made up of the terms of a geometric sequence and is represented by the symbol. You can work with other sorts of series as well, but you won’t have much experience with them until you get to calculus. For the time being, you’ll most likely be collaborating with these two. How to deal with arithmetic series is explained and shown on this page, among other things. You can only take the “partial” sum of an arithmetic series for a variety of reasons that will be explored in greater detail later in calculus.
The following is the formula for the firstnterms of the anarithmeticsequence, starting with i= 1, and it is written: Content Continues Below The “2” on the right-hand side of the “equals” sign may be converted to a one-half multiplied on the parenthesis, which reveals that the formula for the total is, in effect,n times the “average” of the first and final terms, as seen in the example below.
The summation formula may be demonstrated via induction, by the way.
Find the35 th partial sum,S 35, of the arithmetic sequence with terms
The first thirty-five terms of this sequence are added together to provide the 35th partial sum of the series. The first few words in the sequence are as follows: Due to the fact that all of the words share a common difference, this is in fact an arithmetic sequence. The final term in the partial sum will be as follows: Plugging this into the formula, the 35 th partial sum is:Then my answer is:35 th partial sum:Then my answer is:35 th partial sum: S 35 = 350 S 35 = 350 If I had merely looked at the formula for the terms in the series above, I would have seen the common difference in the above sequence.
If we had used a continuous variable, such as the “x” we used when graphing straight lines, rather than a discrete variable, then ” ” would have been a straight line that rose by one-half at each step, rather than the discrete variable.
Find the value of the following summation:
It appears that each term will be two units greater in size than the preceding term based on the formula ” 2 n– 5 ” for the then-thirteenth term. (If I wasn’t sure about something, I could always plug in some values to see if they were correct.) As a result, this is a purely arithmetic sum. However, this summation begins at n= 15, not at n= 1, and the summation formula is only applicable to sums that begin at n=1. So, how am I supposed to proceed with this summation? By employing a simple trick: The quickest way to determine the value of this sum is to first calculate the 14th and 47th partial sums, and then subtract the 14th from the 47th partial sum.
By performing this subtraction, I will have subtracted the first through fourteenth terms from the first through forty-seventh terms, and I will be left with the sum of the fifteenth through forty-seventh terms, as shown in the following table.
These are the fourteenth and forty-seventh terms, respectively, that are required: a14= 2(14) – 5 = 23a47= 2(47) – 5 = 89a14= 2(14) – 5 = 23a47= 2(47) – 5 = 89 With these values, I now have everything I need to find the two partial sums for my subtraction, which are as follows: I got the following result after subtracting: Then here’s what I’d say: As a side note, this subtraction can also be written as ” S 47 – S 14 “.
Don’t be surprised if you come across an exercise that employs this notation and requires you to decipher its meaning before you can proceed with your calculations; this is common.
If you’re dealing with something more complex, however, it may be necessary to group symbols together in order to make the meaning more obvious. In order to do so correctly, the author of the previous exercise should have formatted the summation with grouping symbols in the manner described below:
Find the value ofnfor which the following equation is true:
Knowing that the first term has the value a1= 0.25(1) + 2 = 2.25, I may proceed to the second term. It appears from the formula that each term will be 0.25 units larger than the preceding term, indicating that this is an arithmetical series withd= 0.25, as shown in the diagram. The summation formula for arithmetical series therefore provides me with the following results: The number n is equal to 2.25 + 0.25 + 2 = 42n is equal to 0.25 + 4.25 + 42 = 420.25 n2+ 4.25 n– 42 = 0n2+ 17 n– 168 = 0(n+ 24)(n– 7 = 0n2+ 17 n– 168 = 0(n+ 24)(n– 7).
Then n= 7 is the answer.
However, your instructor may easily assign you a summation that needs you to use, say, eighty-six words or a thousand terms in order to arrive at the correct total.
As a result, be certain that you are able to do the calculations from the formula.
Find the sum of1 + 5 + 9 +. + 49 + 53
After looking through the phrases, I can see that this is, in fact, an arithmetic sequence: The sum of 5 and 1 equals 49 and 5 equals 453 and 49 equals 4. The reason for this is that they won’t always inform me, especially on the exam, what sort of series they’ve given me. (And I want to get into the habit of checking this way.) They’ve provided me the first and last terms of this series, however I’m curious as to how many overall terms there are in this series. This is something I’ll have to sort out for myself.
After plugging these numbers into the algorithm, I can calculate how many terms there are in total: a n=a1+ (n–1) d 53 = 1 + (n–1) a n=a1+ (n–1) (4) 53 = 1 + 4 n– 453 = 4 n– 356 = 4 n– 14 =n 53 = 1 + 4 n– 453 = 4 n– 356 = 4 n– 14 =n There are a total of 14 words in this series.
+ 49 + 53 = 1 + 5 + 9 Then I’ll give you my answer: partial sum S 14 = 378 S 14= 378 After that, we’ll look at geometric series.
Partial Sums of an Arithmetic Sequence
MathJax is the software we utilize. To obtain the sum of an arithmetic series, a finite number of terms from the sequence can be added together. As a result of the unlimited long-term behavior of an arithmetic sequence, we are always limited to adding a finite number of terms to it.
Specific Numerical Results
Take the number$8+13+18+23+ldots+273$ as an example. In a short time, we realize that the terms have a common difference of 5, and that this total is the product of an arithmetic series whose explicit formula is$a n=5n+3$. As a result, the sequence of partial sums is defined by$s n=sumlimits_ n (5k+3)$, where $n$ is any positive integer. When we solve the equation$5n+3=273$, we find out that 273 is the 54th term in the sequence. We might proceed in the following manner if we used a little imagination.
Utilizing the characteristics of sums is an additional method of solving the problem. (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587
Specific Partial Sum Formulas
It is possible to use either strategy to generate a formula for the series of partial sums$s n=sumlimits_ n (5k+3)$, just by leaving $n$ as a variable in the equation. As a result, we get the explicit formulae for the sequence of partial sums that are shown below.
- $s n=dfrac(a 1+a n)=dfrac(8+5n+3)=dfrac $
- _$sumlimits_ _$sumlimits_ _$sumlimits_ n (5k+3)=5sumlimits_ n k+sumlimits_ n 3=(5)dfrac+3n=dfrac n 5=(5)dfrac+3n=dfrac n 3=(5)dfrac+3n=dfrac $
Various applications for the arithmetic sequence may be found on the internet. Examples include the following:
- The seats in an auditorium are arranged in a semicircular configuration, similar to that of a theater. There are 30 rows in total, with the first row having 24 seats and each successive row having an extra 2 seats each row after that. In the explicit formula, the last row will have$a_ =2(30)+22=82$seats, and there will be$s n=dfrac(24+82)=1590$seats in the auditorium
- Campbell County’s grain production was 150 million bushels in 1990, and it has been rising at a rate of 3 million bushels per year since then. With the explicit formula ($a n=3n+147$), the county produced$a_ =3(21)+147=210$million bushels in the year 2010, and a total of$s_ =dfrac(150+210)=3780$million bushels throughout the course of those twenty-one years.
An example of a Partial Sum is the sum of a portion of a sequence.
This is the sequence of even integers starting with 2 and going up to infinity: There are four words in that sequence, and this is the partial sum of those four terms: 2 + 4 + 6 + 8 = 20 Now, let us define the terms a bit more precisely: A sequence is a collection of items (typically numbers) that are arranged in a specific order. A Partial Sum is the sum of a portion of a sequence of numbers. An Infinite Series is made up of the sum of all infinite terms. Partial sums are also referred to as “Finite Series” in some circles.
Partial sums are frequently written with the phrase “add them all up” in mind: This sign (calledSigma) is derived from the Greek word for “sum up,” which implies to bring everything together.
|Thevalues are shown belowand above the Sigma:||4Σn=1n|
|it saysngoes from 1 to 4,which is1,2,3and4|
OK, Let’s Go.
|So now we add up 1,2,3 and 4:||4Σn=1n = 1 + 2 + 3 + 4 =10|
It is common to see partial sums printed with the phrase “add them all up” in the text: So, to summarize, this sign (calledSigma) indicates “add up” or “compile.”
But you have the ability to accomplish far more powerful things than that! We may square each time and add the results to get the following: The number n 2 equals one plus two plus three plus four plus five equals thirty. We can put the first four words in this sequence together to get the total. 2n+1: The sum of (2n+1) is 3 + 5 + 7 + 9 = 24. We may also use other letters; for example, we can useiand sum upi (i+1), going from 1 to 3: the product of i(i+1) = 12 plus 23 plus 34 = 20 Furthermore, we can begin and conclude with any number.
Partial Sums have a number of beneficial qualities that may be used to assist us in doing the computations.
Multiplying by a Constant Property
Suppose we have a summarization we want to make, let’s say we name ita ka kcould be 2 or k(k-7)+2 or anything really, and cis any constant number (like 2, or-9.1, or anything really), then: In other words, if every term we are summing is multiplied by a constant, we may “pull” the constant outside thesigma by multiplying the constant by the total of the terms.
So, let’s say there’s something we want to sum up, and we’ll name ita ka, where ka may be anything from 2 to 7+2, or anything else at all. If andc is a constant value (such as 2, or -9.1, or something else), then: As a result, if every term we are summing is multiplied by a constant, we may “pull” the constant outside of a square root of the total of its squares.
Adding or Subtracting Property
Which means that when two terms are put together and we wish to sum them up, we may actually sum them individually and then add the results together, which is a valuable knowledge.
It will be simpler to do the two sums separately and then combine them at the end.
It should be noted that this also works for subtraction:
Lastly, here are some helpful shortcuts that make the sums a whole lot easier. In each example, we are attempting to sum from 1 to some valuen.
|Summing the constantcequalsctimesn|
|A shortcut when summingk|
|A shortcut when summingk 2|
|A shortcut when summingk 3|
|Also true when summingk 3|
|Summing odd numbers|
Let’s have a look at some of them:
Example 1: You sell concrete blocks for landscaping.
A buyer has expressed interest in purchasing the full “pyramid” of blocks that you keep out front. The stack is 14 blocks in height. What is the total number of blocks in there? Because each layer is a square, the computation goes as follows: 1 2+ 2 2+ 3 2+ 1 2+ 2 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2 However, the following is a far more straightforward way to write it: We may utilize the following formula fork 2 from the previous section: That was a lot less difficult than adding 1 2+ 2 2+ 3 2+.
+ 14 2 to get a total of 1 2+ 2 2+.
Example 2: The customer wants a better price.
According to the client, the bricks on the outside of the pyramid should be less expensive because they need to be cleaned. You acknowledge that you have read and understand the following: What is the overall cost of the project? You may use this formula to figure out how many “inner” and “outer” blocks are in any layer (except the first). As a result, the cost per layer is as follows:
- Cost (outer blocks) = $7 x 4 (size-1)
- Cost (inner blocks) = $11 x (size-2) 2
As a result, all layers (except from the first) will cost: Now that we know the total, let’s see if we can make the computations a little easier! Using the “Addition Property” from the previous section: Using the “Multiply by Constant Property” from before, we can create the following: That’s a nice thing. However, because we are starting from i=2 instead of i=1, we are unable to take any shortcuts. IF, ON THE OTHER HAND, WE INVENT TWO NEW VARIABLES: We have the following:(I removed the k=0 example because I am aware that 0 2 =0) And now we may take advantage of the shortcuts: After a quick computation, we arrive at: $7 364 plus $11 650 equals $9,698.00.
And don’t forget about the top layer (size=1), which is made up of only one single block.
Keep in mind that when we combine the “outer” and “inner” blocks together together with the one on top, we obtain the sum of 364 + 650 + 1 = 1015.
Arithmetic Sequences and Sums
A sequence is a collection of items (typically numbers) that are arranged in a specific order. Each number in the sequence is referred to as aterm (or “element” or “member” in certain cases); for additional information, see Sequences and Series.
An Arithmetic Sequence is characterized by the fact that the difference between one term and the next is a constant. In other words, we just increase the value by the same amount each time. endlessly.
1, 4, 7, 10, 13, 16, 19, 22, and 25 are the numbers 1 through 25.
Each number in this series has a three-digit gap between them. Each time the pattern is repeated, the last number is increased by three, as seen below: As a general rule, we could write an arithmetic series along the lines of
- There are two words: Ais the first term, and dis is the difference between the two terms (sometimes known as the “common difference”).
1, 4, 7, 10, 13, 16, 19, 22, and 25 are the numbers 1 through 25. Has:
- In this equation, A = 1 represents the first term, while d = 3 represents the “common difference” between terms.
And this is what we get:
It is possible to define an Arithmetic Sequence as a rule:x n= a + d(n1) (We use “n1” since it is not used in the first term of the sequence).
Example: Write a rule, and calculate the 9th term, for this Arithmetic Sequence:
3, 8, 13, 18, 23, 28, 33, and 38 are the numbers three, eight, thirteen, and eighteen. Each number in this sequence has a five-point gap between them. The values ofaanddare as follows:
- A = 3 (the first term)
- D = 5 (the “common difference”)
- A = 3 (the first term).
In this equation, A = 3 (the first term), d = 5 (the “common difference”), and e = 3.
Advanced Topic: Summing an Arithmetic Series
To summarize the terms of this arithmetic sequence:a + (a+d) + (a+2d) + (a+3d) + (a+4d) + (a+5d) + (a+6d) + (a+7d) + (a+8d) + (a+9d) + (a+9d) + (a+9d) + (a+9d) + (a+9d) + (a+9d) + ( make use of the following formula: What exactly is that amusing symbol? It is referred to as The Sigma Notation is a type of notation that is used to represent a sigma function. Additionally, the starting and finishing values are displayed below and above it: “Sum upnwherengoes from 1 to 4,” the text states. 10 is the correct answer.
Example: Add up the first 10 terms of the arithmetic sequence:
The values ofa,dandnare as follows:
- A = 1 (which is the first term)
- D = 3 (the “common difference” between two words)
- D = 3 (the “common difference” between two terms)
- N = 10 (number of words to add together)
To summarize, the equation becomes:= 2+9 + 3 = 5(29). Why don’t you try adding the terms yourself and check whether it comes out to 145?
Footnote: Why Does the Formula Work?
Let’s take a look at why the formula works because we’ll be employing an unusual “technique” that’s worth understanding. First, we’ll refer to the entire total as “S”: S = a + (a + d) +. + (a + (n2)d) +(a + (n1)d) + (a + (n2)d) + (a + (n1)d) + (a + (n2)d) + (a + (n1)d) + (a + (n1)d) + (a + (n2)d) + (a + (n1)d) + (a + (n1)d) + (a + (n1)d) + After that, rewrite S in the opposite order: S = (a + (n1)d)+ (a + (n2)d)+. +(a + d)+a. +(a + d)+a. +(a + d)+a. Now, term by phrase, add these two together:
|S||=||a||+||(a+d)||+||.||+||(a + (n-2)d)||+||(a + (n-1)d)|
|S||=||(a + (n-1)d)||+||(a + (n-2)d)||+||.||+||(a + d)||+||a|
|2S||=||(2a + (n-1)d)||+||(2a + (n-1)d)||+||.||+||(2a + (n-1)d)||+||(2a + (n-1)d)|
Each and every term is the same! Furthermore, there are “n” of them. 2S = n (2a + (n1)d) = n (2a + (n1)d) Now, we can simply divide by two to obtain the following result: The function S = (n/2) (2a + (n1)d) is defined as This is the formula we’ve come up with: