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Homesum of odd numbers

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# sum of odd numbers

The sum of the first $n$ positive odd integers can be calculated by using the well-known property of the arithmetic progression, that the sum of its terms is equal to the arithmetic mean of the first and the last term, multiplied by the number of the terms:

$\underbrace{1+3+5+7+9+\cdots+(2n\!-\!1)}_{{n}}=n\cdot\frac{1\!+\!(2n\!-\!1)}{2% }=n^{2}$ |

Thus, the sum of the first $n$ odd numbers is $n^{2}$ (this result has been proved first time in 1575 by Francesco Maurolico).

Below, the odd numbers have been set to form a triangle, each $n^{{\rm{th}}}$ row containing the next $n$ consecutive odd numbers. The arithmetic mean on the row is $n^{2}$ and the sum of its numbers is $n\cdot n^{2}=n^{3}$.

$\displaystyle\begin{array}[]{cccccccccccccccccc}&&&&&&&&&1&&&&&&&&\\ &&&&&&&&3&&5&&&&&&&\\ &&&&&&&7&&9&&11&&&&&&\\ &&&&&&13&&15&&17&&19&&&&&\\ &&&&&21&&23&&25&&27&&29&&&&\\ &&&&31&&33&&35&&37&&39&&41&&&\\ &&&&&\vdots&&&&\vdots&&&&\vdots&&&&\\ \end{array}$ |

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NumberOdd

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## Comments

## sum of even numbers

And adding 1 to each number, in Mr. pahio entry (Hi my friend!), we get the sum of n first even numbers, i.e. n^2+n=n(n+1).

## Geometric proof

There's also the standard geometric 'proof' of this statement that you might want to add. Draw a square matrix of dots. Then the size of the set of dots

\{S_n=(x,y) | x\leq n, y=n or y\leq n, x=n\}

is 2n+1; regarding each of these as a shell, as you add shells, you clearly get the next square.

Roger

## Re: Geometric proof

Roger, you mean the $\lnot$-formed patterns of unit squares which form the n times n square. I think this idea may be a bit too complex for to be added to that entry -- at least I cannot explain it sufficiently clearly. But if you like, feel free and add it.

Jussi