Sum of numbers from 1 to 5. Entertaining mathematics: Gauss rule

Content:

Integers are numbers that do not contain a fractional or decimal part. If the task requires adding a certain number of integers from 1 to a given value N, then they do not need to be added manually. Instead, use the formula (N(N+1))/2, where N is the largest number in the series.

Steps

1 Determine the largest integer (N). By summing the integers from 1 to any given number N, you must determine the value of N (N cannot be a decimal number or a fraction or a negative number).
- Example. Find the sum of all integers from 1 to 100. In this case, N=100, since this is the largest (and final) number of the number series given to you.
2 Multiply N by (N + 1) and divide by 2. When you have determined the integer value N, substitute it into the formula (N(N+1))/2 and you will find the sum of all integers from 1 to N.
- Example. Substitute N=100 and get (100(100+1))/2.
3 Write down the answer. The final answer is the sum of all integers from 1 to the given N.
- Example.
  - (100(100+1))/2 =
  - (100(101))/2 =
  - (10100)/2 = 5050
  - The sum of all integers from 1 to 100 is 5050.
4 Derivation of the formula (N(N+1))/2. Consider the above example again. Mentally divide the row 1 + 2 + 3 + 4 + ... + 99 + 100 into two rows - the first from 1 to 50, and the second from 51 to 100. If you add the first number (1) of the first row and the last number (100 ) of the second row, you get 101. You also get 101 if you add 2 and 99, 3 and 98, 4 and 97, and so on. If each number of the first group is added to the corresponding number of the second group, then in the end we get 50 numbers, each of which is equal to 101. Therefore, 50 * 101 \u003d 5050 is the sum of numbers from 1 to 100. Note that 50 \u003d 100/2 and 101 = 100 + 1. In fact, this is true for the sum of any positive integers: their summation can be broken into two stages with two rows of numbers, and the corresponding numbers in each row can be added to each other, and the result of the addition will be the same.
- We can say that the sum of integers from 1 to N is (N/2)(N+1). A simplified version of this formula is the formula (N(N+1))/2.

Calculating the sum of numbers located between two numbers, using the sum from 1 to N

1 Define the summation option (inclusive or not). Often in tasks, instead of finding the sum of numbers from 1 to a given number N, they are asked to find the sum of integers from N 1 to N 2, where N 2 > N 1 and both numbers > 1. Calculating such a sum is quite simple, but before Before proceeding with the calculations, you must determine whether the given numbers in N 1 and N 2 are included in the final sum or not.
2 To find the sum of integers between two numbers N 1 and N 2 , separately find the sum up to N 1 , separately find the sum up to N 2 and subtract them from each other (subtract the sum up to the smaller N from the sum up to the larger N). In this case, it is important to know whether to sum up inclusively or not. When summing inclusively, you must subtract 1 from the given value N 1 ; otherwise, you must subtract 1 from the given value N 2 .
- Example. Find the sum (“inclusive”) of integers from N 1 = 75 to N 2 = 100. In other words, we must find 75 + 76 + 77 + ... + 99 + 100. To solve the problem, we must find the sum of integers from 1 to N 1 -1, and then subtract it from the sum of numbers from 1 to N 2 (remember: when summing inclusive, we subtract 1 from N 1):
  - (N 2 (N 2 + 1))/2 - ((N 1 -1)((N 1 -1) + 1))/2 =
  - (100(100 + 1))/2 - (74(74 + 1))/2 =
  - 5050 - (74(75))/2 =
  - 5050 - 5550/2 =
  - 5050 - 2775 = 2275. The sum of numbers from 75 to 100 ("inclusive") is 2275.
- Now let's find the sum of the numbers without including the given numbers (in other words, we have to find 76 + 77 + ... + 99). In this case, we subtract 1 from N 2:
  - ((N 2 -1)((N 2 -1) + 1))/2 - (N 1 (N 1 + 1))/2 =
  - (99(99 +1))/2 - (75(75 + 1))/2 =
  - (99(100))/2 - (75(76))/2 =
  - 9900/2 - 5700/2 =
  - 4950 - 2850 \u003d 2100. The sum of numbers from 75 to 100 (without including these numbers) is 2100.
3 Understand the process. Think of the sum of integers from 1 to 100 as 1 + 2 + 3 +... + 98 + 99 + 100 and the sum of integers from 1 to 75 as 1 + 2 + 3 + ... + 73 + 74 + 75. The sum of the integers from 75 to 100 ("inclusive") is the calculation: 75 + 76 + 77 + ... + 99 + 100. The sum of the numbers from 1 to 75 and the sum of the numbers from 1 to 100 are equal to the number 75, but the sum of the numbers from 1 to 100 after the number 75 continues: ... + 76 + 77 + ... + 99 + 100. Thus, subtracting the sum of numbers from 1 to 75 from the sum of numbers from 1 to 100, we “isolate” the sum of integers from 75 to 100.
- If we are summing inclusively, we must use the sum from 1 to 74, not the sum from 1 to 75, to include the number 75 in the final sum.
- Similarly, if we sum without including these numbers, we must use the sum from 1 to 99, not the sum from 1 to 100, to exclude the number 100 from the final sum. We can use the sum from 1 to 75, since subtracting it from the sum from 1 to 99 eliminates the number 75 from the final sum.

The result of the calculation of the sum is always an integer, because either N or N + 1 is an even number that is divisible by 2 without a remainder.
Amount = Amount - Amount.
In other words: Sum = n(n+1)/2

Warnings

Although it is not very difficult to extend this method to negative numbers, this article only considers any positive integers N where N is greater than or equal to 1.

help me please!! calculate the sum of natural numbers from 1+2+3+4+...+97+98+99+100. and got the best answer

Answer from Alexander Heinonen[guru]
The outstanding German mathematician Carl Friedrich Gauss (1777-1855) was called by his contemporaries "the king of mathematics".
Even in early childhood, he showed outstanding mathematical abilities. At the age of three, Gauss was already correcting his father's accounts.
They say that in the elementary school where Gauss (6 years old) studied, the teacher, in order to occupy the class for a long time with independent work, gave the task to the students - to calculate the sum of all natural numbers from 1 to 100. Little Gauss answered the question almost instantly, which is unbelievable surprised everyone and, above all, the teacher.
Let's try to verbally solve the problem of finding the sum of the above numbers. First, let's take the sum of numbers from 1 to 10: 1 + 2 + 3 + 4 + 5 + 6 + +7 + 8 + 9 + 10.
Gauss found that 1 + 10 = 11, and 2 + 9 = 11, and so on. He determined that when adding natural numbers from 1 to 10, 5 such pairs are obtained, and that 5 times 11 is equal to 55.
Gauss saw that the addition of the numbers of the entire series should be carried out in pairs, and he compiled an algorithm for quickly adding numbers from 1 to 100.
1 2 3 4 5 6 7 8 …49 50 51 52 …94 95 96 97 98 99 100
1. It is necessary to count the number of pairs of numbers in the sequence from 1 to 100. We get 50 pairs.
2. Add the first and last numbers of the entire sequence. In our case, these are 1 and 100. We get 101.
3. We multiply the number of pairs of numbers in the sequence by the amount obtained in paragraph 2. We get 5050.
Thus, the sum of natural numbers from 1 to 100 is 5050.
Simple formula: sum of numbers from 1 to n = n * (n+1) : 2. Replace n with the last number and calculate.
Check it out! It works!

Answer from Ianya Fertikova[newbie]
5050

Answer from Mikhail Medvedev[active]
5050

Answer from Pavel Solomennikov[newbie]
5050

Answer from Alevtina bashkova[newbie]
5050

Answer from Ђigr Tikhomirova[active]
5050

Answer from Maria Dubrovina[newbie]
5050

Answer from Aavil Badirov[newbie]
5050

Answer from Dmitry[active]
5050

Answer from Evgeny Sayapov[active]
5050

Answer from 2 answers[guru]

The "Entertaining Mathematics" cycle is dedicated to children who are fond of mathematics and parents who devote time to the development of their children, "throwing" them with interesting and entertaining tasks, puzzles.

The first article in this series is devoted to the Gauss rule.

A bit of history

The famous German mathematician Carl Friedrich Gauss (1777-1855) differed from his peers from early childhood. Despite the fact that he was from a poor family, he learned to read, write, and count quite early. In his biography there is even a mention that at the age of 4-5 years he was able to correct the error in the incorrect calculations of his father, simply by watching him.

One of his first discoveries was made at the age of 6 in a mathematics class. The teacher needed to captivate the children for a long time and he proposed the following problem:

Find the sum of all natural numbers from 1 to 100.

Young Gauss coped with this task quite quickly, having found an interesting pattern, which has become widespread and is still used in mental counting.

Let's try to solve this problem orally. But first, let's take the numbers from 1 to 10:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

Look carefully at this sum and try to guess what was unusual about Gauss? To answer, you need to have a good understanding of the composition of numbers.

Gauss grouped the numbers as follows:

(1+10) + (2+9) + (3+8) + (4+7) + (5+6)

Thus, little Karl received 5 pairs of numbers, each of which individually gives 11 in total. Then, in order to calculate the sum of natural numbers from 1 to 10, you need

Let's return to the original problem. Gauss noticed that before summing, it is necessary to group numbers into pairs, and thereby invented an algorithm thanks to which you can quickly add numbers from 1 to 100:

1 + 2 + 3 + 4 + 5 + … + 48 + 49 + 50 + 51 + 52 + 53 + … + 96 + 97 + 98 + 99 + 100

Find the number of pairs in a series of natural numbers. In this case, there are 50.

Sum the first and last numbers of this series. In our example, these are 1 and 100. We get 101.

We multiply the resulting sum of the first and last member of the series by the number of pairs of this series. We get 101 * 50 = 5050

Therefore, the sum of natural numbers from 1 to 100 is 5050.

Tasks for using the Gauss rule

And now your attention is invited to problems in which the Gauss rule is used to one degree or another. These puzzles are quite capable of being understood and solved by a fourth-grader.

You can give the child the opportunity to reason for himself, so that he himself “invented” this rule. And you can take it apart and see how he can use it. Among the tasks below are examples in which you need to understand how to modify the Gauss rule in order to apply it to a given sequence.

In any case, in order for the child to operate with this in his calculations, it is necessary to understand the Gaussian algorithm, that is, the ability to divide correctly into pairs and count.

Important! If a formula is memorized without understanding, then it will be forgotten very quickly.

Task 1

Find the sum of numbers:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10;
1 + 2 + 3 + … + 14 + 15 + 16;
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9;
1 + 2 + 3 + 4 + 5 + … + 48 + 49 + 50 + 51 + 52 + 53 + … + 96 + 97 + 98 + 99 + 100.

Solution.

At first, you can give the child the opportunity to solve the first example himself and offer to find a way in which it is easy to do it in the mind. Next, analyze this example with the child and show how Gauss did it. For clarity, it is best to write down a series and connect pairs of numbers with lines that add up to the same number. It is important that the child understands how pairs are formed - we take the smallest and largest of the remaining numbers, provided that the number of numbers in the row is even.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = (1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6) = (1 + 10) * 5;
1 + 2 + 3 + … + 14 + 15 + 16 = (1 + 16) + (2 + 15) + (3 + 14) + (4 + 13) + (5 + 12) + (6 + 11) + (7 + 10) + (8 + 9) = (1 + 16) * 8 = 136;
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = (1 + 8) + (2 + 7) + (3 + 6) + (4 + 5) + 9 = (1+ 8) * 4 + 9 = 45;
1 + 2 + 3 + 4 + 5 + … + 48 + 49 + 50 + 51 + 52 + 53 + … + 96 + 97 + 98 + 99 + 100 = (1 + 100) * 50 = 5050

A task2

There are 9 weights weighing 1g, 2g, 3g, 4g, 5g, 6g, 7g, 8g, 9g. Can these weights be divided into three piles of equal weight?

Solution.

Using the Gauss rule, we find the sum of all weights:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = (1 + 8) * 4 + 9 = 45 (g)

So, if we can group the weights so that each pile contains weights with a total weight of 15g, then the problem is solved.

One of the options:

9g, 6g
8g, 7g
5g, 4g, 3g, 2g, 1g

Find other possible options yourself with your child.

Pay attention to the child that when such problems are solved, it is better to always start grouping with a larger weight (number).

Task 3

Is it possible to divide the clock face into two parts by a straight line so that the sums of the numbers in each part are equal?

Solution.

To begin with, apply the Gauss rule to the series of numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12: find the sum and see if it is divisible by 2:

So you can share. Now let's see how.

Therefore, it is necessary to draw a line on the dial so that 3 pairs fall into one half, and three into the other.

Answer: the line will pass between the numbers 3 and 4, and then between the numbers 9 and 10.

A task4

Is it possible to draw two straight lines on the clock face so that the sum of the numbers in each part is the same?

Solution.

To begin with, we apply the Gauss rule to the series of numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12: find the sum and see if it is divisible by 3:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = (1 + 12) * 6 = 78

78 is divisible by 3 without a remainder, so you can divide. Now let's see how.

According to the Gauss rule, we get 6 pairs of numbers, each of which adds up to 13:

1 and 12, 2 and 11, 3 and 10, 4 and 9, 5 and 8, 6 and 7.

Therefore, it is necessary to draw lines on the dial so that 2 pairs fall into each part.

Answer: the first line will pass between the numbers 2 and 3, and then between the numbers 10 and 11; the second line is between the numbers 4 and 5, and then between 8 and 9.

Task 5

A flock of birds is flying. Ahead is one bird (leader), followed by two, then three, four, etc. How many birds are in the flock if there are 20 of them in the last row?

Solution.

We get that we need to add numbers from 1 to 20. And to calculate such a sum, we can apply the Gauss rule:

1 + 2 + 3 + 4 + 5 + … + 15 + 16 + 17 + 18 + 19 + 20 = (20 + 1) * 10 = 210.

Task 6

How to seat 45 rabbits in 9 cages so that all cages have a different number of rabbits?

Solution.

If the child has decided and understood the examples from task 1 with understanding, then it is immediately remembered that 45 is the sum of numbers from 1 to 9. Therefore, we put the rabbits like this:

first cell - 1,
second - 2,
third - 3,
eighth - 8,
ninth - 9.

But if the child cannot figure it out right away, then try to give him the idea that such problems can be solved by brute force and you need to start with the minimum number.

Task 7

Calculate the sum using the Gauss trick:

31 + 32 + 33 + … + 40;
5 + 10 + 15 + 20 + … + 100;
91 + 81 + … + 21 + 11 + 1;
1 + 2 + 3 + 4 + … + 18 + 19 + 20;
1 + 2 + 3 + 4 + 5 + 6;
4 + 6 + 8 + 10 + 12 + 14;
4 + 6 + 8 + 10 + 12;
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11.

Solution.

31 + 32 + 33 + … + 40 = (31 + 40) * 5 = 355;
5 + 10 + 15 + 20 + … + 100 = (5 + 100) * 10 = 1050;
91 + 81 + … + 21 + 11 + 1 = (91 + 1) * 5 = 460;
1 + 2 + 3 + 4 + … + 18 + 19 + 20 = (1 + 20) * 10 =210;
1 + 2 + 3 + 4 + 5 + 6 = (1 + 6) * 3 = 21;
4 + 6 + 8 + 10 + 12 + 14 = (4 + 14) * 3 = 54;
4 + 6 + 8 + 10 + 12 = (4 + 10) * 2 + 12 = 40;
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = (1 + 10) * 5 + 11 = 66.

Task 8

There is a set of 12 weights weighing 1g, 2g, 3g, 4g, 5g, 6g, 7g, 8g, 9g, 10g, 11g, 12g. 4 weights were removed from the set, the total mass of which is equal to a third of the total mass of the entire set of weights. Can the remaining weights be placed on two balance pans, 4 pieces on each pan, so that they are in balance?

Solution.

We apply the Gauss rule to find the total mass of the weights:

1 + 2 + 3 + ... + 10 + 11 + 12 = (1 + 12) * 6 = 78 (g)

We calculate the mass of weights that have been removed:

Therefore, the remaining weights (with a total mass of 78-26 \u003d 52 g) must be placed 26 g on each scale pan so that they are in balance.

We don't know which weights have been removed, so we have to consider all possible options.

Applying the Gauss rule, you can divide the weights into 6 pairs with equal weight (13g each):

1g and 12g, 2g and 11g, 3g and 10, 4g and 9g, 5g and 8g, 6g and 7g.

Then the best option is when removing 4 weights, two pairs of the above will be removed. In this case, we will have 4 pairs left: 2 pairs on one scale and 2 pairs on the other.

The worst case is when 4 weights removed will break 4 pairs. We will have 2 unbroken pairs with a total weight of 26g, which means we place them on one scale pan, and the remaining weights can be placed on another scale pan and they will also be 26g.

Good luck with the development of your children.