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 Hi Maths Central My wife presented me with a query which may have a simple answer, but one that I can’t deduce or explain. Take the string of numbers, n=1,2,3,4,5… It seems that n(2n-1) = Sum((1…..(2n-1)) e.g. for n=5, both 5 x 9 and Sum(1….9) equal 45, and so on for other values of n. Could you please provide an explanation? Does it have an underlying reason and a name? Look forward to your response. Brian

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

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