Find a quick and easy method to compute the sum of the first 100 counting numbers. I can't use a calculator. I can't figure out an easy way.
1+2+3+4+5...........98+99+100
Carl Freidrich Gauss had no idea as to how famous his name would become when he solved a seemingly time consuming arithmetic problem posed by his teacher with amazing speed. At age 10, his teacher presented his class with a long problem in addition, the answer to which he could find by formula in a few seconds. Some historical references indicate that the problem was of the following type, 72365 + 72550 + 72735 + .........+ 90865, where the common difference was 185, and the total number of terms was 100. Other sources claim that the class was asked to add up the first hundred whole numbers. As the story goes, it was customary at the school for the first person to get the answer to lay his paper or slate face down on the table. The teacher had barely finished presenting the problem when Gauss, alledgedly, placed his slate on the table, and sat quietly for the rest of the hour while his classmates struggled. Upon reviewing the slates at the end of the period, he found but a single number on Gauss's slate, the correct answer. Gauss supposedly did not know the trick for solving such arithmetic progression problems so quickly and his teacher was quite surprised.
History records that Gauss created the method instantaneously in his head. It is said that he did it by pairing the terms and then multiplying the value of each pair by the number of pairs. Taking the simpler problem of the first 100 integers to illustrate, the pairs all total 100, i.e., 100+0 = 100, 99+1 = 100, 98+2 = 100, .........51+49 = 100. This results in 50 pairs of 100 each totalling 5000, plus the 50 left over in the middle, for a total of 5050.
Other historical records say that he imagined the sum he sought, denoted by S, as being written in both ascending and descending order:
S = 1 + 2 + 3 + 4 + .....................................98 + 99 + 100
S = 100 + 99 + 98 + 97...........................................3 + 2 + 1
Now, instead of adding the numbers horizontally across the rows, Gauss supposedly added them vertically down the columns. In so doing, he derived S = 101 + 101 + 101 + 101 + ...............................101 + 101 + 101
as the sum of each column is simply 101. There being 100 columns, it was quickly obvious to Gauss that 2S = 100 x 101 = 10100 and therefore the sum of the first hundred whole numbers became S = 1 + 2 + 3 + 4 + .................+ 98 + 99 + 100 = 100(101)/2 = 10100/2 = 5050.
Considering the alledgedly more difficult problem posed by the teacher, and the fact that he supposedly wrote but a single number on his slate, I guess it would have been possible for Gauss to mentally add 72365 and 90865, multiply by 100, and divide by 2, giving him his single answer of 8,161,500. Obviously, the 1 through 100 version is a bit more believable. Either way, his teacher considered it a stroke of pure genius.
It is worth mentioning that any book on number theory will offer you the expressions for the sum of the first odd numbers 1 through n as (1+3+5+7+....+(2n-1) = n^2 and the sum of the first even numbers 2 through 2n as S(2+4+6+....(2n) = n(n-1). Therefore, if we add all the odd and even integers from 1 through n, we will get
S = (n/2)^2 + (n/2)(n/2 + 1) = n^2/4 + n^2/4 + n/2 = n^2/2 + n/2 = n(n + 1)/2
What do you know? Look familiar? S(1 + 100) = 100(101)/2 = 5050.
There are a few, more familiar, ways to derive the answer.
1--You can simply take the average of the first and last numbers, or (1+100)/2 = 50.5 and multiply it by 100 giving you 5050.
2--The traditional formal approach, is to view it as an arithmetic progression with the first term a = 1, the common difference d = 1, the last term l = 100, and the number of terms n = 100. The sum of an arithmetic progression is given by S = n(a+l)/2 which, for the 1-100 case, becomes S = 100(1+100)/2 = 5050 again. (Note that this
equation is merely the mathematical expression of what we did in the first case.)
3--You might also recognize that the sum of the 10 digits from 1 to 10 is 55 and adding the next set of 10 digits merely adds another 55 plus 10 x 10 = 100 for a total of 155, the next set totaling 255, the next 355, and so on up to 955. The sum of this sequence is then 10(55) + 9(100+900)/2 = 550 + 4500 = 5050.
There are several methods you can use to compute the sum of the first 100 counting numbers (1+2+3+4+5...+98+99+100) without using a calculator.
Method 1: Pairing and multiplying
This method is based on the approach used by Carl Friedrich Gauss to solve the problem quickly. You can pair up the numbers from both ends of the sequence, where the sum of each pair is equal to 100. For example, 1+99=100, 2+98=100, 3+97=100, and so on. Since there are 50 pairs, you can multiply 100 by 50 to get 5000. Additionally, you can add the remaining number in the middle, which is 50. Therefore, the sum of the first 100 counting numbers is 5000 + 50 = 5050.
Method 2: Ascending and descending order
In this method, you can visualize the sum of the counting numbers as being written in both ascending and descending order. The sum can be represented as follows:
S = 1+2+3+4+...+98+99+100
S = 100+99+98+97+...+3+2+1
Now, instead of adding the numbers horizontally, you can add them vertically down the columns. In each column, the sum is simply 101 since there are 100 columns. Therefore, the total sum is 101 multiplied by 100, which equals 10100. However, each number has been counted twice (once in the ascending order and once in the descending order). So, you divide 10100 by 2 to get the sum of the first 100 counting numbers, which is 5050.
Method 3: Formula for the sum of an arithmetic progression
You can use the formula for the sum of an arithmetic progression to compute the sum of the first 100 counting numbers. In this case, the first term is 1, the common difference is 1, and the number of terms is 100. The formula for the sum of an arithmetic progression is S = n(a + l)/2, where S is the sum, n is the number of terms, a is the first term, and l is the last term. Plugging in the values, the equation becomes S = 100(1 + 100)/2, which simplifies to S = 5050.
Method 4: Summing in groups
You can also sum the numbers in groups. For example, you can recognize that the first 10 numbers (1+2+3+...+8+9+10) sum up to 55. Then, for the next set of 10 numbers (11+12+13+...+18+19+20), you add 55 to 10 times 10 (since there are 10 groups of 10). This pattern continues until you reach the last set of 10 numbers (91+92+93+...+98+99+100), where you add 55 to 9 times 10 (since there are 9 groups of 10 at the end). By using this approach, the sum becomes 10(55) + 9(100+900)/2 = 550 + 4500 = 5050.
All these methods yield the same result, which is 5050 for the sum of the first 100 counting numbers.