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Tuesday, September 30, 2003 - 11:58 AM
If you want to add all the numbers from F (First) to L (Last), here is an easy way to do it, and many times can be performed in your head.∑(F,L)= ( L² - F² + F + L)/2
Where: F is the first number in the number line and L is the Last.
Example:
To add all the numbers from 1 to 10 Plug in 1 for F, and 10 for L.
(10² - 1² + 10 + 1)/2 =
(100 - 1 + 11)/2 =
110/2 = 55
The result is the same as adding all the numbers from 1 to 10 like this:
1+2+3+4+5+6+7+8+9+10=55
This works for all nonnegative integers where 0<F<L. You don't have to start with 1.
Where: F is the first number in the number line and L is the Last.
Example:
To add all the numbers from 1 to 10 Plug in 1 for F, and 10 for L.
(10² - 1² + 10 + 1)/2 =
(100 - 1 + 11)/2 =
110/2 = 55
The result is the same as adding all the numbers from 1 to 10 like this:
1+2+3+4+5+6+7+8+9+10=55
This works for all nonnegative integers where 0<F<L. You don't have to start with 1.
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Comments
1 2 3 4 5
10 9 8 7 6
The you add them vertically, and they all becomes 11. You have 5 of these sums --> total of 55. This can as the previous example be done with all numbers.
I think it was Euler who first did this in an excersise at school, when the teacher wanted time for more interesting things. He simply asked the students to add all numbers between 1 and 100. He thought that he would have plenty of time for his own research, but Euler(?) solved it within a minute with the method mentioned above.
Find the middle number and then multiply by the number of numbers.
eg. 7+8+9+10+11 =9 * 5 = 45
eg. 7+8+9+10+11+12 = 9.5 * 6 = 57
(Trick here: there are 6 numbers so there is no proper middle one. Use the number half way between the TWO middle ones)
I had heard the story and that is what inspired me to come up with my formula, which I have named my Gaussian Summation in his honor.
As a point of interest, my formula was originally born in my head as a polynomial. I had to use the standard FOIL method to convert it to the final, more useful form you see above.
I like your solution! Thanks very much for the input! -- E.J. Wilson
Thank you and best wishes! - E.J. Wilson
Method:
This is a modification to the method suggested by Wramde on Oct 15, 2003 - 01:45 AM.
It seems eassier to compute if we consider the following algebraic form of the same:
(F+L)*N/2
where F is the first term, L is the last term and N is the number of numbers in the sequence.
-Joe Philip Ninan
Title : Connecting all three methods suggested by Deud, Wramde and Clay
We can derive all the three methods from one
Method:-
The method suggested by Wramde on Oct 15, 2003 - 01:45 AM. can be written as
(F+L)*N/2
N is the number of numbers from F to L. ie, N = (L - F)+1
Therefore (F+L)*N/2 = (F+L)*[(L - F)+1] / 2
When we simplify we will get
( L² - F² + F + L)/2
This is what Deud suggested on Sep 30, 2003 - 08:58 AM.
Back to Wramde's method.
(F+L)*N/2 =
=(F+L)/2 *N.
This is what Clay suggested on Nov 04, 2003 - 07:12 AM
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Another formula I would like to suggest is
N/2 * [2F + (N - 1)]
Here we don't need to know the last number.
take the example ,sum of numbers 1-10.
1) take highest number 10/2=5
2) to highest number 10 add 1= 11
3)multiply 11*5=55=ans
example:-sum of numbers from 6-10
1) take highest number 10/2=5
2) take highest number 10 + 1=11
3) multiply last 2 ans ie; 11*5=55
4) take numbers 1-5(missing from 1-10(leaving 6-10 above))
5) take highest number 5/2=2.5
6)take highest number 5+1=6
7)multiply last 2 ans ie;6*2.5=15
8)take ans from (3)minus ansfrom (7)=40
9) so ans to sum of numbers from 6-10 is:-
55-15=40
looks complicated ,but easier to do than read!! lol
i dont think it was euler,i heard it was fredriech gauss (if i spelt it right!lol),but i could be wrong!
∑(F,L)= ( L² - F² + L + F) / 2
NOT
∑(F,L)= ( L² - F² + F + L)/2
okay...just keeping an eye out!
We can use it to add all even (or odd) numbers from F to L:
2+4+6+8+10=30
or
(2+10)*5/2=30
---------------------------------
The only catch is that the numbers need to be contstant increments (Add the same number to the previous number) of each other. Example:
1+4+7+10+13+16 = 51
(1+16)*6/2 = 51
---------------------------------
This next example does not work because we are multipling instead of adding:
1+3+9+27 = 40
(1+27)*4/2 != 40
Example: what is the sum of the numbers 1 through 10.
1) take highest number and multiply it by highest number plus 1. 10*11=110.
2) divide answer by 2. 110/2= 55
I have heard the exact same story about Niels Henrik Abel... It may just be a myth?
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