A. Gauss and the Derivation of a Formula

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A. Gauss and the Derivation of a Formula

Carl Friedrich Gauss was a brilliant mathematician born in Germany in 1777. When he was a child (around age 10), his teacher posed the question, “What is the sum of the numbers from \(1\) to \(100\)?” The rumor is that this was a tedious problem posed by the teacher to busy the students, since adding \(1\) to \(100\) would take quite a while to do! The teacher was shocked when Carl Gauss was able to give the correct answer within a few minutes.

How did he do it so fast?

To start off, he wrote out the sum forward and backward, and then added pairs of numbers, as shown below. What do you notice when you do this?
How many sums of \(101\) are there?

However, the result of adding these \(100\) sets together would be double the desired sum (remember that the sum from \(1\) to \(100\) was written twice). Therefore, to find the sum of the numbers from \(1\) to \(100\) ONCE, multiply \(100\) by \(101\), and then divide the result by \(2\).

\[1 + 2 + 3 + ... + 98 + 99 + 100 = \frac{{100 \cdot 101}}{2} = 5050 \]

Without question, this is a much more efficient method of finding the sum than adding each number separately!

How is this going to help find a formula for the sum of an arithmetic series?

First, note that when \(1, 2, 3, …, 100\) is written using commas, it is an arithmetic sequence. The first term is \(1\) and the common difference is \(1\). Written with addition symbols, the arithmetic sequence becomes an arithmetic series! In this case, \(1 + 2 + 3 + … + 100\) is an arithmetic series.

Suppose you were asked to find the sum of a general arithmetic series. The sum, \(S_n \) ,would be written

\(t_1  + \left[ {t_1  + d} \right] + \left[ {t_1  + 2d} \right] + \left[ {t_1  + 3d} \right] + ... + \left[ {t_1  + (n - 2)d} \right] + \left[ {t_1  + (n - 1)d} \right] \)

Following Gauss’ example, add the sum forward and backward together, and see if you can find a pattern (which leads to a formula)!
\[\begin{align}
 S_n &= t_1 & + & (t_1 + d) & + ... + & (t_1 + (n - 2)d) & + (t_1 + (n - 1)d) \\
 +S_n &= (t_1 + (n - 1)d) & + & (t_1 + (n - 2)d) & + ... + & (t_1 + d) & + t_1 \\
\hline {} \\ 2S_n &= (2t_1 + (n - 1)d) & + & (2t_1 + (n - 1)d) & + ... + & (2t_1 + (n - 1)d) & + (2t_1 + (n - 1)d) \\ \end{align} \]

A pattern appears; each pair adds to \(2t_1 + (n - 1)d \). How many pairs are there? The number of pairs is equal to the number of terms, \(n \). To finish it off, multiply the number of terms, \(n \), by the sum of each pair, \(2t_1 + (n - 1)d \), and then divide by \(2\) because you originally doubled the sum.
\[
S_n  = \frac{{n \cdot \left[ {2t_1  + (n - 1)d} \right]}}{2} = \frac{n}{2}\left[ {2t_1  + (n - 1)d} \right]
\]