



 
Hi Brian, A very nice observation by your wife! Her is one way to deduce the expression she observed. Let \[S = 1 + 2 + 3 + \cdot + \cdot + \cdot + (2n  3) + (2n  2) + (2n  1)\] I am going to write the expression for $S$ again starting at $(2n  1)$ and decreasing to $1$ and write the two expressions under each other. \begin{eqnarray*} Now I am going to add the two expressions together to get $2S$ adding downwards \begin{eqnarray*} There are $ (2n  1)$ terms on the right side so \[2S = (2n  1) \times 2n\] or \[S = n (2n  1).\] A sequence where the difference between successive terms is a constant is called an Arithmetic Progression. The difference between successive terms is called the Common Difference. You have the arithmetic progression that starts at 1, has a common difference of 1 and has $(2n  1)$ terms. The technique above is one of the ways to find the sum of an arithmetic progression. For example you could use this technique to find \[17 + 20 + 23 + \cdot \cdot \cdot + 80.\] Penny  


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