# sum of odd numbers

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

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

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  $n\cdot n^{2}=n^{3}$.

 $\displaystyle\begin{array}[]{cccccccccccccccccc}&&&&&&&&&1&&&&&&&&\\ &&&&&&&&3&&5&&&&&&&\\ &&&&&&&7&&9&&11&&&&&&\\ &&&&&&13&&15&&17&&19&&&&&\\ &&&&&21&&23&&25&&27&&29&&&&\\ &&&&31&&33&&35&&37&&39&&41&&&\\ &&&&&\vdots&&&&\vdots&&&&\vdots&&&&\\ \end{array}$
Title sum of odd numbers SumOfOddNumbers 2013-03-22 14:38:35 2013-03-22 14:38:35 pahio (2872) pahio (2872) 15 pahio (2872) Example msc 00A05 msc 11B25 NumberOdd