“Add up all of the numbers between 1 and 100.”

Easy, once you realize that (1, 100) make a pair whose sum is 101, (2, 99) make another such pair, etc.

“Add up all of the numbers between 1 and 101.”

Harder to make sense of, because there’s an odd number of integers that you’re summing, so they can’t be paired off evenly.

I’ve got (at least) pieces of student work that show (at least) 3 different ways of handling an odd number of integers, as in the problem above.

In the comments below, try to catch ’em all. I’ll update this post with the actual student work tomorrow.

Update: For pictures of student work on this, see the next day’s post.

  • Method 1: Especially if these two questions were posed in this order, I think the easiest to do would be:
    (1 + 2+ 3 + … + 100) + 101. That is, take the previous answer and add 101.

    Method 2: A related (but I think much harder way) would be to find out which is the “middle” number that won’t get paired in the way that you show and take it out, then sum the remaining, then add back in that middle number.

    Method 3: Another way, I show my students how:
    S = 1 + 2 + 3 + … + 101
    S = 101 + 100 + 99 + … + 1

    Then sum the two lines column by column to get:
    2S = 102 + 102 + 102 + … + 102 = 101 * 102
    S = (101 * 102)/2

    In a little more generality, that is how I prove the Finite Arithmetic Series formula.

  • First method and add the extra number.

  • They can use a formula like s = (n+1) * n/2. It’s interesting to me what this formula represents when you use an odd number of values.

  • Instead of “sum of n+1 times n/2 pairs” you can think of it as “average of (n+1)/2 times n numbers” so that you don’t have to worry about whether n is even or odd.