ydoaPs Posted December 16, 2006 Share Posted December 16, 2006 I was playing around with my calculator during class the other day. (Yes, I still don't pay attention in class.) In doing so, I discovered something. [math]n^2=1+2+3+...+(n-1)+n+(n-1)+...+3+2+1[/math] Is there a name for this? Link to comment Share on other sites More sharing options...

woelen Posted December 16, 2006 Share Posted December 16, 2006 This is fairly basic math. In fact, if you sum up powers of i, with i ranging from 1 to n, then you'll see that summing a k-th power yields a number of k+1th power (plus some correction terms of lower power). This can be connected to integrating functions. If you take the integral of x^k, then the result has power k+1. In your case, you can connect it to integrating x, which gives an expression of order x^2. Link to comment Share on other sites More sharing options...

ydoaPs Posted December 16, 2006 Author Share Posted December 16, 2006 So, no name? Link to comment Share on other sites More sharing options...

EvoN1020v Posted December 17, 2006 Share Posted December 17, 2006 I'm thinking perhaps "factorial"? i.e. 5! = 5x4x3x2x1 n(n-1)(n-2)(n-3)...1 But it probably doesn't fit with your pattern because you are adding the terms. It looks like if a person is starting at the bottom of a mountain, and it's going uphill. Then at the top of the peak, it goes downhill back to the bottom. Anyways. Link to comment Share on other sites More sharing options...

uncool Posted December 17, 2006 Share Posted December 17, 2006 It would usually be done wth mathematical induction - each time, you add 2n + 1, making the numbe ron the left side n^2 + 2n + 1 = (n+1)^2, and as you have a base case with 1 working, mathematical induction guarantees it for all n. =Uncool- Link to comment Share on other sites More sharing options...

The Thing Posted December 19, 2006 Share Posted December 19, 2006 Arithmetic Series. Get the value of [math]1+2+3+...+n[/math], add it to [math](n-1)+(n-2)+...+1[/math]: [math]\frac{n(n+1)}{2}+\frac{n(n-1)}{2}[/math] Simplify, gives the series as [math]n^2[/math]. Link to comment Share on other sites More sharing options...

John Cuthber Posted December 20, 2006 Share Posted December 20, 2006 Or, if you prefer a geometric equivalent, the 2 halves of the expression on the right are 2 triangles, put together with a line of n "stars" they form a square. * *** ** ** *** * **** Can someone who understood that and who has a better graphics package than me draw that please? Link to comment Share on other sites More sharing options...

The Thing Posted December 21, 2006 Share Posted December 21, 2006 Hey, that's quite nice! Put the series in pairs like this: [math]1+n-1+2+n-2+...+n-1+1+n[/math] [math]=n+n+n...+n[/math]. There are [math]n[/math] n's adding together. Which is the square that John drew above. So: [math]n^2[/math] is the series's sum. Link to comment Share on other sites More sharing options...

ssd Posted February 4, 2007 Share Posted February 4, 2007 Hey, that's quite nice! Put the series in pairs like this: [math]1+n-1+2+n-2+...+n-1+1+n[/math] [math]=n+n+n...+n[/math]. There are [math]n[/math] n's adding together. Which is the square that John drew above. So: [math]n^2[/math] is the series's sum. Very nice and quite emphatic. Link to comment Share on other sites More sharing options...

alex Posted February 23, 2007 Share Posted February 23, 2007 cool Link to comment Share on other sites More sharing options...

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