Finding the sum of a series.

[AviaryPhoenix]

Hi, I am quite new to the Turing software. I have several codes for sequences and counting. However, I need to write a program that will total all the numbers from 1 to 100.
Any help is appreciated. Thank You

[Tony]

code:

put "5050"

What happened to the template form for new posts?

[AviaryPhoenix]

Tony @ Sun Feb 27, 2011 7:06 pm wrote:

code:

put "5050"

What happened to the template form for new posts?

Oh was i supposed to keep that?

Anyways I know the sum of the series from 1-100 is 5050, but is there any program that would total the series, no matter what numbers are used?

[Tony]

Yes, there is such a program. You should be able to write it. You could ask Carl Friedrich Gauss for advice, if you want to make it super fast.

[mirhagk]

I love you tony. LOL.

Guass is honestly my favourite mathematician ever.

I believe the story goes that his teacher wanted to punish him by forcing him to add all the numbers from 1 to 100 by hand. He realized that if you take 100 and 1 and add them it's 101. 2 and 99 also equal 101, same for 3 and 98, 4 and 97 etc...

So he took the first and last element and added them together 50 times (since you use 2 numbers each time). So it works out to (first number+last number)*number of numbers/2

There is also a mathematical proof proving this requiring surprisingly little algabra if you are interested.

[A.J]

There are two proofs actually, one involving telescoping the series (n+1)^2 - n^2, and the other involving writing the numbers in increasing and decreasing order to show Gauss' method. However, telescoping is a useful method that you can use to find the closed form of the sum of the first n powers of k.

So, to get the sum of the first n numbers, the telescoping series method would be as follows:

Consider the following summation:

sum(k = 1 -> n) [(k+1)^2 - k^2] = (n+1)^2 - 1 (since everything except for (n+1)^2 and -1^2 cancels).

However, (k+1)^2 - k^2 = 2k + 1. So:

sum (k = 1 -> n) [2k + 1] = (n+1)^2 - 1, or:

2*sum(k = 1 -> n) [k] + n = (n+1)^2 - 1, and thus sum(k = 1 -> n) [k] = n*(n+1)/2

So, in general, considering the summation sum(k = 1 -> n) [(k+1)^r - k^r] will yield the closed form for the sum of the first n powers r.

But to answer your original question, AviaryPhoenix, to compute the sum of the first n numbers, you either use the closed form derived above, or you try iterating through the numbers from 1 to n using a for loop, adding the values to some variable.