Need Help with an Algebra(?) Formula

(not homework…mine or anyone elses)

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 :slight_smile: )

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.

Is there a formula to easy calculate the result?

K * (K+1) / 2

For example when K = 5, 1+2+3+4+5 =
5 * 6 / 2

(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.)

Here: https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_⋯

Or – for the sum of 1:N use the formula k = n(n+1) / 2

For example for 1:5 –

k = 5 * (6) / 2 = 30/2 = 15 = 1 + 2 + 3 + 4 + 5

G’luck!

Thank you!

If you really want to have a lot saved up by Christmas, try putting aside $1 the first week, $2 the second week, $4 in Week 3, $8 in Week 4, etc.

:wink:

You’re EVIL.

God bless us, every one!

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.

Okay, That’s pretty cool. I’ve never thought of it that way.

4.5 quadrillion later

Yes! If every atom in the universe was a printing press, each churning out dollar bills at the rate of atomic vibrations since the Big Bang . . . :smiley:

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.

Another trick if you forget the formula – Just add the first to the last, the second to the second last and so on.

For this case:
(1 + 52) + (2 + 51) + (3 + 50) + … =
53 + 53 + 53 + … =
26 * 53
(There will be half the total number of terms, or 26 in this case.)

Its easy to do it in excel…

But in mathematics its called an arithmetic series.

I seem to recall " K! " Where K, in this instance, would be 52.

No, that’s if he was multiplying.

In this case, the quick way to do it has already been given but the formula for each value in the series would be a summatory:

n
Ʃ x
x=1

That means “add all values of x from 1 to n”; n would be the number of each week, from 1 to 52.

$1388, if I understand the question correctly.

Close. 26 X $53 = $1378.

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.

Speaking of factorials, I heard a good factorial puzzle years ago.

The value of 6! is 720, which ends in one trailing zero.

The value of 12! is 479,001,600 which ends in two trailing zeroes.

How many trailing zeroes does 2016! end with?