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 Question from Tim, a student: How do I develop a rule for the sum of all whole numbers from 1 to X when I have no idea how to do this

Hi Tim,

Lets start with an example and then move to the general case.

How do you add $1 + 2 + 3 + \cdot \cdot \cdot + 48 + 49 + 50?$ To do this I suggest that you start by adding from both ends and work toward the middle.

$1 + 50 = 51$

$2 + 49 = 51$

$3 + 48 = 51$

and so on.

when you get to the middle you have

$25 + 26 = 51$

Hence you have $51, 25$ times so

$1 + 2 + 3 + \cdot \cdot \cdot + 48 + 49 + 50 = 51 \times 25.$

In words what I see is that $1 + 2 + 3 + \cdot \cdot \cdot + 48 + 49 + 50$ is the first term plus the last term, times the number of terms divided by 2.

There is a complication if the number of terms is odd since in that case when you start adding from both ends toward the middle there is an extra term in the middle. There is a cleaver way to show that the rule above applies in all cases so let's do it for the general case.

Suppose $S = 1 + 2 + 3 + \cdot \cdot \cdot + (X - 2)+ (X - 1)+ X.$ Now write $S$ again starting from the right end, $S = X + (X - 1) + (X - 2) + \cdot \cdot \cdot + 3 + 2 + 1.$ Thus I have

$S = 1 + 2 + 3 + \cdot \cdot \cdot + (X - 2)+ (X - 1)+ X$

$S = X + (X - 1) + (X - 2) + \cdot \cdot \cdot + 3 + 2 + 1$

$2S = (X + 1) + (X + 1) + (X + 1) + \cdot \cdot \cdot + (X + 1) + (X + 1) + (X + 1)$

Hence $2S$ is $(X + 1)$ times the number of terms in the original sequence and thus $S = X + (X - 1) + (X - 2) + \cdot \cdot \cdot + 3 + 2 + 1$ is the first term plus the last term, times the number of terms divided by 2.

Your textbook probably has a formula for this using $a, n, \mbox{ and } d$ but I don't remember formulas well. This technique is valid for any arithmetic sequence, that is a sequence where the difference between any tow consecutive terms is always the same. Thus for example you can use it to add $2 + 4 + 6 + \cdot \cdot \cdot + 98 + 100$ or $3 + 6 + 9 + \cdot \cdot \cdot + 99$ or even $5 + 8 + 11 + \cdot \cdot \cdot + 65.$

I hope this helps,
Penny

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