Sum of odd numbers results in a square number - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-22T17:35:05Z http://mathoverflow.net/feeds/question/4348 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4348/sum-of-odd-numbers-results-in-a-square-number Sum of odd numbers results in a square number LiraNuna 2009-11-06T08:15:07Z 2009-11-06T16:30:01Z <p>Recently I discovered by myself that sum of N sequential odd numbers will result into N<sup>2</sup>.</p> <p>Can anyone explain this to me? I want to know <em>why</em> not a proof.</p> <p>Some examples: </p> <pre> 1 1<sup>2</sup> 1 1+3 2<sup>2</sup> 4 1+3+5 3<sup>2</sup> 9 ... 1+3+5+7+9+11+13+15 8<sup>2</sup> 64 </pre> <p>This is not homework, but personal knowledge. I'm also sorry in advance if it's too simple.<br /> Also an apology if it's tagged wrong.</p> http://mathoverflow.net/questions/4348/sum-of-odd-numbers-results-in-a-square-number/4351#4351 Answer by Darsh Ranjan for Sum of odd numbers results in a square number Darsh Ranjan 2009-11-06T08:31:20Z 2009-11-06T08:31:20Z <p>Draw an $n \times n$ square:</p> <pre>xxxx xxxx xxxx xxxx</pre> (here $n$ = 4, and there are 16 x's). Now divide the square into $n$ symmetric L-shapes: <pre> dcba dcbb dccc dddd</pre> <p>As you can see, we have 1 a, 3 b's, 5 c's and 7 d's, so 16 = 1 + 3 + 5 + 7. </p> http://mathoverflow.net/questions/4348/sum-of-odd-numbers-results-in-a-square-number/4354#4354 Answer by Aaron Mazel-Gee for Sum of odd numbers results in a square number Aaron Mazel-Gee 2009-11-06T08:39:28Z 2009-11-06T08:39:28Z <p>Another way of seeing this is to compare two consecutive square numbers, say n<sup>2</sup> and (n+1)<sup>2</sup>. If we expand the larger one, we get (n+1)<sup>2</sup>=n<sup>2</sup>+2n+1, so it is exactly (2n+1) more than the previous square. As n increases, this expression represents the consecutive odd numbers. So squares always differ by consecutive odd numbers.</p> http://mathoverflow.net/questions/4348/sum-of-odd-numbers-results-in-a-square-number/4360#4360 Answer by Jose Brox for Sum of odd numbers results in a square number Jose Brox 2009-11-06T09:40:15Z 2009-11-06T16:30:01Z <p>I know you didn't ask for a proof, but actually the best way of seeing WHY something happens is understanding a good, simple proof for it. Here is my try to your nice observation, using the formula for the sum of several consecutive terms of an arithmetic progression (which, by the way, has also a nice, simple proof):</p> <p>You know that if $\{an+b\}$ is an arithmetic progression and you look at some of its consecutive terms, then their sum is "(the first one plus the last one) times (the number of terms added) divided by 2".</p> <p>In your case, you have the progression $\{2n-1\}$, starting at $1$ and finishing in some $2N-1$, which has $N$ terms. Then, by the formula above, you get that their sum is $$S = \frac{(1+2N-1)\cdot N}{2} = \frac{2N\cdot N}{2} = N^2.$$</p> <p>This, if you look at it by the reverse side, is caused by the fact that the substraction of two consecutive squares is a particular odd number, $(N+1)^2 - N^2 = N^2+2N+1-N^2 = 2N+1$. You could also use this to prove your claim by induction, but it wouldn't be, in my opinion, a clarifying proof of the kind you were looking for.</p>