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*}
S &=& 1 &+& 2 &+& 3 &+& \cdot \cdot \cdot &+& (2n-3) &+& (2n-2) &+& (2n-1)\\
S &=& (2n-1) &+& (2n-2) &+& (2n-3) &+& \cdot \cdot \cdot &+& 3 &+& 2 &+& 1
\end{eqnarray*}

Now I am going to add the two expressions together to get $2S$ adding downwards

\begin{eqnarray*}
2S &=& 2n &+& 2n &+& 2n &+& \cdot \cdot \cdot &+& 2n &+& 2n &+& 2n.
\end{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