So, I’m doing that Christmas savings thing where you put away a dollar amount each week based on what week of the year it is.
IOW week 1, you put away $1, week 2, $2, Week 3, $3, and so on to week 52.
What I’ve been doing though is, whenever I have a few extra bucks in my pocket, I’ll stick in the envelope. Or, if for example it’s week 3 and I have a $5 bill, I’ll just put the whole $5 and mark off week 3.
Then every once in a while I’ll count the money and verify what week I’m paid up to. (we’re in the 9th week of the year and I’ve saved through week 20 already )
I imagine there must be some formula for easily figuring out how much $ should have been saved by a certain week.
But, for the life of me, my 47 year-old brain can’t reach far enough back to my 8th grade algebra class to figure it out.
So, I’m stuck with getting out the calculator and punching in “1+2+3+4+5…” to find the answer.
(There’s a famous story that an old schoolmaster asked his misbehaving class to sum the numbers 1 + 2 + 3 + … + 100. The very young Carl Gauss produced the answer in a few seconds.)
Besides memorizing the formula, think of it like you’re filling up a triangle, like this:
**
Now what is the area of that five-row triangle? Notice that if you double it and put the two triangles together, you’d get a rectangle that is five by six.
ooooo
*oooo
**ooo
***oo
****o
And the area of a five by six rectangle is 30, so one five-row triangle is 15. It’s easy to generalize this, the sum of the first n integers is n times (n+1), all divided by 2.
As the story is told (at least as I heard it from my 4th grade teacher), this isn’t how Gauss did it.
Gauss’s thought process, as I heard it told, went like this: He noticed that 1+99=100. And 2+98=100. And 3+97=100. And . . . up to 49+51=100. So that’s a total of 49 hundreds. And the final number, 100 itself makes 50 hundreds. And the number 50 (not yet accounted) makes 5050.
That’s close to how I did it. I did 1 + 100 = 101. 2 + 99 = 101. Etc. That way I was able to multiply 101 X 50 = 5050. It was when I contemplated a series with an odd number of members (leaving one unpairable number in the center that I realized that I needed a universal formula, and came up with [x (× + 1)]/2.
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