# The Formula Gives The Partial Sum Of An Arithmetic Sequence. What Is The Formula Solved For An? (Best solution)

The formula s=n(a1+an)/2 gives the partial sum of an aritmetic sequence.

## What is the formula for the partial sum of an arithmetic sequence?

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.

## What is the formula for the arithmetic sequence if the sum of the same sequence is given by SN 2n 3n?

Sn=2n + 3n?? an= 5 + 6(n-1)

## How will you solve for the sum of the arithmetic sequence formed?

The Sum Formula The formula says that the sum of the first n terms of our arithmetic sequence is equal to n divided by 2 times the sum of twice the beginning term, a, and the product of d, the common difference, and n minus 1. The n stands for the number of terms we are adding together.

## What is the formula for arithmetic series?

An arithmetic sequence can be defined by an explicit formula in which an = d (n – 1) + c, where d is the common difference between consecutive terms, and c = a1.

## 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?

A partial sum of an infinite series is the sum of a finite number of consecutive terms beginning with the first term. When working with infinite series, it is often helpful to examine the behavior of the partial sums.

n(a+b)

## What is the common difference of the AP if the sum of n terms of an AP is given by SN 3n 2n 2?

If the sum of n terms of an A.P. is given by Sn = 3n + 2n2, then the common difference of the A.P. is 4.

## What is the formula of sum of n terms?

We use the first term (a), the common difference (d), and the total number of terms (n) in the AP to find its sum. The formula used to find the sum of n terms of an arithmetic sequence is n/2 (2a+(n−1)d).

## Arithmetic Series

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 more in size than the preceding term based on the formula ” 2 n– 5 ” for the then-thirteenth term. (Whether 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 simplest approach to get 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 doing 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 total of the fifteenth through forty-seventh terms, as shown in the following table.

These are the fourteenth and forty-seventh words, respectively, that are required: a14= 2(14) – 5 = 23a47= 2(47) – 5 = 89a14= 2(14) – 5 = 23a47= 2(47) – 5 = 89 With these numbers, I now have everything I need to get 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 may also be written as ” S 47 – S 14 “.

Don’t be shocked if you come into an exercise that use this notation and requires you to decipher its meaning before you can proceed with your calculations; this is common.

If you’re working with anything more complicated, though, it may be important 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 structured the summation using grouping symbols in the manner shown 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).

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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.

## Arithmetic Sequences and Series

The succession of arithmetic operations There is a series of integers in which each subsequent number is equal to the sum of the preceding number and specified constants. orarithmetic progression is a type of progression in which numbers are added together. This term is used to describe a series of integers in which each subsequent number is the sum of the preceding number and a certain number of constants (e.g., 1). an=an−1+d Sequence of Arithmetic Operations Furthermore, becauseanan1=d, the constant is referred to as the common difference.

For example, the series of positive odd integers is an arithmetic sequence, consisting of the numbers 1, 3, 5, 7, 9, and so on.

This word may be constructed using the generic terman=an1+2where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, where, To formulate the following equation in general terms, given the initial terma1of an arithmetic series and its common differenced, we may write: a2=a1+da3=a2+d=(a1+d)+d=a1+2da4=a3+d=(a1+2d)+d=a1+3da5=a4+d=(a1+3d)+d=a1+4d⋮ As a result, we can see that each arithmetic sequence may be expressed as follows in terms of its initial element, common difference, and index: an=a1+(n−1)d Sequence of Arithmetic Operations In fact, every generic word that is linear defines an arithmetic sequence in its simplest definition.

### Example 1

Identify the general term of the above arithmetic sequence and use that equation to determine the series’s 100th term. For example: 7,10,13,16,19,… Solution: The first step is to determine the common difference, which is d=10 7=3. It is important to note that the difference between any two consecutive phrases is three. The series is, in fact, an arithmetic progression, with a1=7 and d=3. an=a1+(n1)d=7+(n1)3=7+3n3=3n+4 and an=a1+(n1)d=7+(n1)3=7+3n3=3n+4 and an=a1+(n1)d=3 As a result, we may express the general terman=3n+4 as an equation.

To determine the 100th term, use the following equation: a100=3(100)+4=304 Answer_an=3n+4;a100=304 It is possible that the common difference of an arithmetic series be negative.

### Example 2

Identify the general term of the given arithmetic sequence and use it to determine the 75th term of the series: 6,4,2,0,−2,… Solution: Make a start by determining the common difference, d = 4 6=2. Next, determine the formula for the general term, wherea1=6andd=2 are the variables. an=a1+(n−1)d=6+(n−1)⋅(−2)=6−2n+2=8−2n As a result, an=8nand the 75thterm may be determined as follows: an=8nand the 75thterm a75=8−2(75)=8−150=−142 Answer_an=8−2n;a100=−142 The terms in an arithmetic sequence that occur between two provided terms are referred to as arithmetic means.

### Example 3

Find all of the words that fall between a1=8 and a7=10. in the context of an arithmetic series Or, to put it another way, locate all of the arithmetic means between the 1st and 7th terms. Solution: Begin by identifying the points of commonality. In this situation, we are provided with the first and seventh terms, respectively: an=a1+(n−1) d Make use of n=7.a7=a1+(71)da7=a1+6da7=a1+6d Substitutea1=−8anda7=10 into the preceding equation, and then solve for the common differenced result. 10=−8+6d18=6d3=d Following that, utilize the first terma1=8.

a1=3(1)−11=3−11=−8a2=3 (2)−11=6−11=−5a3=3 (3)−11=9−11=−2a4=3 (4)−11=12−11=1a5=3 (5)−11=15−11=4a6=3 (6)−11=18−11=7} In arithmetic, a7=3(7)11=21=10 means a7=3(7)11=10 Answer: 5, 2, 1, 4, 7, and 8.

### Example 4

Find the general term of an arithmetic series with a3=1 and a10=48 as the first and last terms. Solution: We’ll need a1 and d in order to come up with a formula for the general term. Using the information provided, it is possible to construct a linear system using these variables as variables. andan=a1+(n−1) d:{a3=a1+(3−1)da10=a1+(10−1)d⇒ {−1=a1+2d48=a1+9d Make use of a3=1. Make use of a10=48. Multiplying the first equation by one and adding the result to the second equation will eliminate a1.

an=a1+(n−1)d=−15+(n−1)⋅7=−15+7n−7=−22+7n Answer_an=7n−22 Take a look at this! Identify the general term of the above arithmetic sequence and use that equation to determine the series’s 100th term. For example: 32,2,52,3,72,… Answer_an=12n+1;a100=51

## Arithmetic Series

Series of mathematical operations When an arithmetic sequence is added together, the result is called the sum of its terms (or the sum of its terms and numbers). Consider the following sequence: S5=n=15(2n1)=++++= 1+3+5+7+9+25=25, where S5=n=15(2n1)=++++ = 1+3+5+7+9=25, where S5=n=15(2n1)=++++= 1+3+5+7+9 = 25. Adding 5 positive odd numbers together, like we have done previously, is manageable and straightforward. Consider, on the other hand, adding the first 100 positive odd numbers. This would be quite time-consuming.

When we write this series in reverse, we get Sn=an+(and)+(an2d)+.+a1 as a result.

2.:Sn=n(a1+an) 2 Calculate the sum of the first 100 terms of the sequence defined byan=2n1 by using this formula.

The sum of the two variables, S100, is 100 (1 + 100)2 = 100(1 + 199)2.

### Example 5

The sum of the first 50 terms of the following sequence: 4, 9, 14, 19, 24,. is to be found. The solution is to determine whether or not there is a common difference between the concepts that have been provided. d=9−4=5 It is important to note that the difference between any two consecutive phrases is 5. The series is, in fact, an arithmetic progression, and we may writean=a1+(n1)d=4+(n1)5=4+5n5=5n1 as an anagram of the sequence. As a result, the broad phrase isan=5n1 is used. For this sequence, we need the 1st and 50th terms to compute the 50thpartial sum of the series: a1=4a50=5(50)−1=249 Then, using the formula, find the partial sum of the given arithmetic sequence that is 50th in length.

### Example 6

Evaluate:Σn=135(10−4n). This problem asks us to find the sum of the first 35 terms of an arithmetic series with a general terman=104n. The solution is as follows: This may be used to determine the 1 stand for the 35th period. a1=10−4(1)=6a35=10−4(35)=−130 Then, using the formula, find out what the 35th partial sum will be. Sn=n(a1+an)2S35=35⋅(a1+a35)2=352=35(−124)2=−2,170 2,170 is the answer.

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### Example 7

In an outdoor amphitheater, the first row of seating comprises 26 seats, the second row contains 28 seats, the third row contains 30 seats, and so on and so forth. Is there a maximum capacity for seating in the theater if there are 18 rows of seats? The Roman Theater (Fig. 9.2) (Wikipedia) Solution: To begin, discover a formula that may be used to calculate the number of seats in each given row. In this case, the number of seats in each row is organized into a sequence: 26,28,30,… It is important to note that the difference between any two consecutive words is 2.

where a1=26 and d=2.

As a result, the number of seats in each row may be calculated using the formulaan=2n+24.

In order to do this, we require the following 18 thterms: a1=26a18=2(18)+24=60 This may be used to calculate the 18th partial sum, which is calculated as follows: Sn=n(a1+an)2S18=18⋅(a1+a18)2=18(26+60) 2=9(86)=774 There are a total of 774 seats available.

Take a look at this! Calculate the sum of the first 60 terms of the following sequence of numbers: 5, 0, 5, 10, 15,. are all possible combinations. Answer_S60=−8,550

### Key Takeaways

• When the difference between successive terms is constant, a series is called an arithmetic sequence. According to the following formula, the general term of an arithmetic series may be represented as the sum of its initial term, common differenced term, and indexnumber, as follows: an=a1+(n−1)d
• An arithmetic series is the sum of the terms of an arithmetic sequence
• An arithmetic sequence is the sum of the terms of an arithmetic series
• As a result, the partial sum of an arithmetic series may be computed using the first and final terms in the following manner: Sn=n(a1+an)2

### Topic Exercises

1. Given the first term and common difference of an arithmetic series, write the first five terms of the sequence. Calculate the general term for the following numbers: a1=5
2. D=3
3. A1=12
4. D=2
5. A1=15
6. D=5
7. A1=7
8. D=4
9. D=1
10. A1=23
11. D=13
12. A 1=1
13. D=12
14. A1=54
15. D=14
16. A1=1.8
17. D=0.6
18. A1=4.3
19. D=2.1
1. Find a formula for the general term based on the arithmetic sequence and apply it to get the 100 th term based on the series. 0.8, 2, 3.2, 4.4, 5.6,.
2. 4.4, 7.5, 13.7, 16.8,.
3. 3, 8, 13, 18, 23,.
4. 3, 7, 11, 15, 19,.
5. 6, 14, 22, 30, 38,.
6. 5, 10, 15, 20, 25,.
7. 2, 4, 6, 8, 10,.
8. 12,52,92,132,.
9. 13, 23, 53,83,.
10. 14,12,54,2,114,. Find the positive odd integer that is 50th
11. Find the positive even integer that is 50th
12. Find the 40 th term in the sequence that consists of every other positive odd integer in the following format: 1, 5, 9, 13,.
13. Find the 40th term in the sequence that consists of every other positive even integer: 1, 5, 9, 13,.
14. Find the 40th term in the sequence that consists of every other positive even integer: 2, 6, 10, 14,.
15. 2, 6, 10, 14,. What number is the term 355 in the arithmetic sequence 15, 5, 5, 15, 25,.
16. What number is the phrase 172 in the arithmetic sequence 4, 4, 12, 20, 28,.
17. What number is the term 355 in the arithmetic sequence 15, 5, 5, 15, 25,.
18. Find an equation that yields the general term in terms of a1 and the common differenced given the arithmetic sequence described by the recurrence relationan=an1+5wherea1=2 andn1 and the common differenced
19. Find an equation that yields the general term in terms ofa1and the common differenced, given the arithmetic sequence described by the recurrence relationan=an1-9wherea1=4 andn1
20. This is the problem.
1. Calculate a formula for the general term based on the terms of an arithmetic sequence: a1=6anda7=42
2. A1=12anda12=6
3. A1=19anda26=56
4. A1=9anda31=141
5. A1=16anda10=376
6. A1=54anda11=654
7. A3=6anda26=40
8. A3=16andananda15=
1. Find all possible arithmetic means between the given terms: a1=3anda6=17
2. A1=5anda5=7
3. A2=4anda8=7
4. A5=12anda9=72
5. A5=15anda7=21
6. A6=4anda11=1
7. A7=4anda11=1

### Part B: Arithmetic Series

1. Make a calculation for the provided total based on the formula for the general term an=3n+5
2. S100
3. An=5n11
4. An=12n
5. S70
6. An=132n
7. S120
8. An=12n34
9. S20
10. An=n35
11. S150
12. An=455n
13. S65
14. An=2n48
15. S95
16. An=4.41.6n
17. S75
18. An=6.5n3.3
19. S67
20. An=3n+5
1. Consider the following values: n=1160(3n)
2. N=1121(2n)
3. N=1250(4n3)
4. N=1120(2n+12)
5. N=170(198n)
6. N=1220(5n)
7. N=160(5212n)
8. N=151(38n+14)
9. N=1120(1.5n2.6)
10. N=1175(0.2n1.6)
11. The total of all 200 positive integers is found by counting them up. To solve this problem, find the sum of the first 400 positive integers.
1. The generic term for a sequence of positive odd integers is denoted byan=2n1 and is defined as follows: Furthermore, the generic phrase for a sequence of positive even integers is denoted by the number an=2n. Look for the following: The sum of the first 50 positive odd integers
2. The sum of the first 200 positive odd integers
3. The sum of the first 50 positive even integers
4. The sum of the first 200 positive even integers
5. The sum of the first 100 positive even integers
6. The sum of the firstk positive odd integers
7. The sum of the firstk positive odd integers the sum of the firstk positive even integers
8. The sum of the firstk positive odd integers
9. There are eight seats in the front row of a tiny theater, which is the standard configuration. Following that, each row contains three additional seats than the one before it. How many total seats are there in the theater if there are 12 rows of seats? In an outdoor amphitheater, the first row of seating comprises 42 seats, the second row contains 44 seats, the third row contains 46 seats, and so on and so forth. When there are 22 rows, how many people can fit in the theater’s entire seating capacity? The number of bricks in a triangle stack are as follows: 37 bricks on the bottom row, 34 bricks on the second row and so on, ending with one brick on the top row. What is the total number of bricks in the stack
10. Each succeeding row of a triangle stack of bricks contains one fewer brick, until there is just one brick remaining on the top of the stack. Given a total of 210 bricks in the stack, how many rows does the stack have? A wage contract with a 10-year term pays $65,000 in the first year, with a$3,200 raise for each consecutive year after. Calculate the entire salary obligation over a ten-year period (see Figure 1). In accordance with the hour, a clock tower knocks its bell a specified number of times. The clock strikes once at one o’clock, twice at two o’clock, and so on until twelve o’clock. A day’s worth of time is represented by the number of times the clock tower’s bell rings.

### Part C: Discussion Board

1. Is the Fibonacci sequence an arithmetic series or a geometric sequence? How to explain: Using the formula for the then th partial sum of an arithmetic sequenceSn=n(a1+an)2and the formula for the general terman=a1+(n1)dto derive a new formula for the then th partial sum of an arithmetic sequenceSn=n2, we can derive the formula for the then th partial sum of an arithmetic sequenceSn=n2. How would this formula be beneficial in the given situation? Explain with the use of an example of your own creation
2. Discuss strategies for computing sums in situations when the index does not begin with one. For example, n=1535(3n+4)=1,659
3. N=1535(3n+4)=1,659
4. Carl Friedrich Gauss is the subject of a well-known tale about his misbehaving in school. As a punishment, his instructor assigned him the chore of adding the first 100 integers to his list of disciplinary actions. According to folklore, young Gauss replied accurately within seconds of being asked. The question is, what is the solution, and how do you believe he was able to come up with the figure so quickly?

1. 5, 8, 11, 14, 17
2. An=3n+2
3. 15, 10, 5, 0, 0
4. An=205n
5. 12,32,52,72,92
6. An=n12
7. 1,12, 0,12, 1
8. An=3212n
9. 1.8, 2.4, 3, 3.6, 4.2
10. An=0.6n+1.2
11. An=6n3
12. A100=597
13. An=14n
14. A100=399
15. An=5n
16. A100=500
17. An=2n32
1. 2,450, 90, 7,800, 4,230, 38,640, 124,750, 18,550, 765, 10,000, 20,100, 2,500, 2,550, K2, 294 seats, 247 bricks, $794,000, and You might be interested: Which Arithmetic Operations Can Be Performed On Pointers? (Perfect answer) ## 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. ### 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 $### Applications Various applications for the arithmetic sequence may be found on the internet. Examples include the following: • The seats in an auditorium are arranged in a semicircular configuration, similar to that of a theater. There are 30 rows in total, with the first row having 24 seats and each successive row having an extra 2 seats each row after that. In the explicit formula, the last row will have$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. ## Partial Sums An example of a Partial Sum is the sum of a portion of a sequence. ### Example: 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. ## Sigma 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. ### Sum What? 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 Here’s how it’s shown in a single diagram: ## More Powerful 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. Here’s how we move from 3 to 5: ## Properties 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. ### Example: Consequently, rather than summing 6k 2, we may sumk 2and then multiply the entire result by 6. ### Adding or Subtracting Property We can sumk 2 instead of summing 6k 2and then multiply the total result by 6 instead. ### Example: 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: ## Useful Shortcuts And here are some essential shortcuts that will make your life a whole lot simpler while doing the math. To summarize, we are attempting the sum from 1 to some valuen in each scenario.  Summing1equalsn 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. This is the same figure we received for the “total blocks” the last time we checked. yay!