Thursday

April 17, 2014

April 17, 2014

Posted by **Tony** on Tuesday, October 3, 2006 at 9:50pm.

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 Some historical references indicate that the problem was of the following type, 72365 + 72550 + 72735 + .........+ 90865, where the common difference was 185, and the total number of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.

**Related Questions**

math - "Find a quick and easy method to compute the sum of the first 100 ...

sum fun (math) - find a quick and easy method to compute the sum of the first ...

Geometry - What is an equation I can use to find the sum of the first 1000 ...

algebra - how do you find the square root of any number with long division and ...

Problem Solving Math Question - Find the sum of all the numbers in the following...

calc - Use L’Hopital’s rule to find the limit of this sequence (n^100)/(e^n) ......

calc2 - Use L’Hopital’s rule to find the limit of this sequence (n^100)/(e^n...

calc - Use L’Hopital’s rule to find the limit of this sequence (n^100)/(e^n) ......

histroy - . When the diplomats met to negotiate the treaty ending the ...

algebra - The sum of the swuares of the first n counting numbers: 1² + 2² + 3...