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What is sum of the first 100 whole numbers?
It is instructive to note that you can rewrite 1+2+3+4....+99+100 as 1+100+2+99+3+98....+50+51 Then you can group each pair of numbers as (1+100)+(2+99)+(3+98)+...+(50+51) All of these couplets add up to 101, and there are 50 of them. Thus the sum of the first 100 whole numbers is 50x101. This can be calculated at 5,050.
When did Carl Friedrich Gauss invent the arithmetic progression?
Actually he did not invent arithmetic progression, but he had this insight as a 7 years old young student. When his teacher asked the class to sum all numbers from 1 to 100, the young Gauss did not need more than a few seconds to write "5050" in his slate. he noticed that 1+100=101, 2+99=101, 3+98=101, ... formed a sequence of 50 pairs that could summarize the calculation to 50x101= 5050. Gauss is today considered by many as the greatest mathematician that ever lived.
What is the sum of the first one hundred counting number?
Here is a nice way to see and remember this. Look at the sum of the first 5 counting number Write them as 1,2,3,4,5 Now write them backward as 5,4,3,2,1 Place the second list under the first list and add the numbers in each column. Each sum is 6 and there are 5 sums. However, you counted the numbers twice since you wrote the list both ways. So the sum of the first 5 numbers is 5x6/2 In general the sum of the first n natural numbers is n(n+1)/2 So the sum of the first 100 numbers is 100x101/2=50x101=5050 This formula can also be proved by induction and several other ways.
What is the mean of the integers from 2 to 100?
First we find the sum of the integers from 2 to 100 including 2 and 100, then we divide by 99 since there are 99 numbers. There are 100 numbers between 1 and 100 and we are excluding only the number 1. If we want to exclude the number 2, and count only the numbers 3,4,5....100, this can be done with the same procedure and a slight modification. So the sum of the numbers 1 to 100 can be found by writing the numbers 1, 2,3,...100 Now write them backwards, starting at 100,99,98....1 Each column has a sum of 101 and there are 100 columns. So the total is 100x101, but we wrote the list twice so we must divide by 2 The sum is 100x101/2=50x101=5050 Now remember we want only 2 to 100 so the sum we seek is 5049. Since there are 99 number, the mean is 5049/99=51 In general the sum of the first n positive integers is n(n+1)/2 This can be proved the way we did or by induction.
How did Karl Gauss sum the integers of 1 to 100?
IN his head, bitchez! Or in longer words: by noting 1+100=101, 2+99=101, ... , 50+51=101 50 pairs of numbers summing to 101, so 50x101 = 5050
What is sum of the first 100 whole numbers?
It is instructive to note that you can rewrite 1+2+3+4....+99+100 as 1+100+2+99+3+98....+50+51 Then you can group each pair of numbers as (1+100)+(2+99)+(3+98)+...+(50+51) All of these couplets add up to 101, and there are 50 of them. Thus the sum of the first 100 whole numbers is 50x101. This can be calculated at 5,050.
When did Carl Friedrich Gauss invent the arithmetic progression?
Actually he did not invent arithmetic progression, but he had this insight as a 7 years old young student. When his teacher asked the class to sum all numbers from 1 to 100, the young Gauss did not need more than a few seconds to write "5050" in his slate. he noticed that 1+100=101, 2+99=101, 3+98=101, ... formed a sequence of 50 pairs that could summarize the calculation to 50x101= 5050. Gauss is today considered by many as the greatest mathematician that ever lived.
What is the sum of the first one hundred counting number?
Here is a nice way to see and remember this. Look at the sum of the first 5 counting number Write them as 1,2,3,4,5 Now write them backward as 5,4,3,2,1 Place the second list under the first list and add the numbers in each column. Each sum is 6 and there are 5 sums. However, you counted the numbers twice since you wrote the list both ways. So the sum of the first 5 numbers is 5x6/2 In general the sum of the first n natural numbers is n(n+1)/2 So the sum of the first 100 numbers is 100x101/2=50x101=5050 This formula can also be proved by induction and several other ways.
What is the mean of the integers from 2 to 100?
First we find the sum of the integers from 2 to 100 including 2 and 100, then we divide by 99 since there are 99 numbers. There are 100 numbers between 1 and 100 and we are excluding only the number 1. If we want to exclude the number 2, and count only the numbers 3,4,5....100, this can be done with the same procedure and a slight modification. So the sum of the numbers 1 to 100 can be found by writing the numbers 1, 2,3,...100 Now write them backwards, starting at 100,99,98....1 Each column has a sum of 101 and there are 100 columns. So the total is 100x101, but we wrote the list twice so we must divide by 2 The sum is 100x101/2=50x101=5050 Now remember we want only 2 to 100 so the sum we seek is 5049. Since there are 99 number, the mean is 5049/99=51 In general the sum of the first n positive integers is n(n+1)/2 This can be proved the way we did or by induction.
