The Easy Way to Add Up a Sequence

[@gawainthedane3314]

At the end I was scared he was going to whip out that skillshare/brilliant sponsorship

[@felpshehe]

Eeeeeeeeeee

[@themexis]

@@felpshehe casual advocacy is casual af

[@Andlekin]

I don't really mind edutainment channels being sponsored by for-profit educational companies.

[@LuizFernando-ek9mh]

That one OG math video you actually understand the logic behind the math

[@NarutoOrganisation13]

"And E..." [Ecstasy slides into view] This must have been a very pleasant video to make, then.

[@somewony]

Oh, that's Ecstasy? For some reason I thought it was an estrogen pill.

[@r15u5k00]

@@somewony so pure :,)

[@NarutoOrganisation13]

@@somewony Haha, I thought so too but I recognized the heart stamp on the pill. Kind of gives it away.

[@TristanBomber]

@@somewony Shit, I thought it was estrogen too, since "E" is slang for estrogen in the trans community. It is pride month, after all!

[@Gillsing]

I'm too innocent to recognise E. Also a bit slow on the uptake, as it took me a moment to figure out that those were all mathematical constants. I thought he was making a rebus.

[@whynotanyting]

As a basement dweller, I accept your apology.

[@r15u5k00]

sound effects guy should get an award

[@Quroe_]

I can always count on you to have the best sponsorships. I never skip 'em.

[@vanillaannihilation5871]

Also the best sound effects.

[@TheBlessingo]

The production value is amazing! I love your work

Love the sound effects while doing calculations!

[@trulyUnAssuming]

Nicely swept the uneven numbers under the table :-p Just in case someone is interested: For uneven n, 0,...,n are n+1 numbers and n+1 is even. So you get (n+1)/2 pairs with value n, which adds up to (n+1)*n/2, which is the same formula as for the even numbers, where the n/2 pairs add up to n+1. And since you can add the numbers up to some n, the formula for adding up numbers between two numbers is just subtracting the ones from 1 which you don't want to add. I.e. the numbers between m and n are the numbers up to n minus the numbers up to m-1. So you get (n+1)*n/2 - m*(m-1)/2=((n+1)*n-m*(m-1))/2

[@AusReddit]

This method works for sequences going by two's as well (2,4,6,8...). Just divide the (last number) - (first number) by another 2 and you've figured it out. As a Mathcounts competitor, knowing this method was a MUST. Either that or you had some extremely fast calculators.

[@haomakk]

ok, now THIS is the best channel on KZbin

[@martinshoosterman]

I don't know why I just watched this video. I'm a 4th year math student at university and I've known about this since the 10th grade. (Not because I figured it out, but because numberphile made a video about this like 8 years ago) But I love this result and this is one of the most well made videos about this result that ive seen.

[@sorenkair]

Didn't mention the equation works with both even and odd number of numbers.

[@ericathefae]

Yeah, I was wondering that - would have loved to hear the explanation for odd numbered lists...

[@ericathefae]

@@Errenium Thank you!

[@icedragonaftermath]

Well, the unpaired number in a sequence of an odd length would have to be halfway between the highest and the lowest values. So the unpaired number should be half * (highest + lowest). The original equation takes that into account though.

[@tommyzheng387]

Caleb Johnsen yeah and there’s a pattern

[@hoomanvassef6483]

Another way to work it out visually/intuitively is to think of the numbers forming a digital "triangle", e.g. (0) + 1 + 2 + 3 + 4 where the numbers are represented by os: xxxx oxxx ooxx ooox oooo But you note the xes also form the same "triangle", and the 2 combined form a rectangle of size 4 x 5, i.e. n [last number] x (n + 1). Hence 1 + 2 + ... + n = n(n+1)/2. For the more general case m + (m+1) + ... + n, the numbers form a digital "trapezoid", made up of one n x (n - m + 1) rectangle, with a digital "triangle" removed from it that's 1/2 of an (n - m) x (n - m + 1) rectangle. => m + (m+1) + ... + n = (n - n/2 + m/2)(n - m + 1) = (n + m)(n - m + 1)/2

[@RedrunLoL]

I'm loving the more frequent uploads

[@Gregorsnek]

this aged poorly...

[@ITR]

You can also do (last + last^2 + first - first^2)/2

[@deidara_8598]

The key to making observations like Gauss did is to think more abstractly about the problem you're solving, rather than use pre-meditaded formulas and equations you've memorized.

[@genessab]

I don’t know why but I find your content so insanely hilarious. I keep laughing wildly on the train

[@hurktang]

Note that it's the mean value multiplied by the number of items. This is how calculus work.

[@Vaaaaadim]

I think that would be more a statistics thing

[@PopeGoliath]

This episode has been brought to you by: Drugs! Yay!

[@jasonpeng5798]

These videos are so satisfying and interesting. They are good to pass time and also I feel like I'm not totally wasting my time. Please make more.

[@gabrielcavuquila4380]

This is actually geometric progression but in a simpler way and I loved it, it would help a lot of my colleagues back in 10th grade

[@ReasonMakes]

Actually the best KZbinr there is

[@Goblin4Coin]

Every video you upload is a delight

[@PeterLiuIsBeast]

If you think about it, it's also quite like the formula for the area of a trapezoid (or triangle with top side = 0).

[@imaytag]

Jesse where did you gooooooooo......?

[@nicolasyan1613]

Good mnemonic/derivation for the sum from 1 to n: make a list of rows with 1, 2, 3, ... n dots in each: (illustrated for n = 4) o oo ooo oooo Then, add on the same sequence in reverse to make a rectangle: o xxxx oo xxx ooo xx oooo x Since it's a rectangle, the number of dots must be n(n+1) (the number of rows is n, the number of columns is n+1). Since you added two of the sequences you wanted to count up, you have to divide by two. So 1+2+...+n = n(n+1)/2

[@videogyar2]

I realized this when I was around 13. It was a similar a question in a regional math test and I was so proud to come up with it lol

[@maxgamesst1]

And then they clapped

[@macskasbogre133]

@@maxgamesst1 It's not a farfetched claim...

[@driveasandwich6734]

To think there was a time were kids could just be the discoverers of mathematical properties by just fiddling around...

[@thewarmedic2330]

As a UIL student I appreciate it a LOT

[@Axian_empire]

Thank god. This video helped me survive my Asian parents

[@lindhe]

You get extra points for the sound effects.

[@tomoyahiro7036]

I just came here to figure out how he explained the method even though I have a prior knowledge about the use of arithmetic series (sum of the arithmetic sequence) formula.

[@Cat_in_Spacetime]

Credits - Jesse Agar

[@scrubby2]

i'm glad this works on odd amount of numbers too.

[@Anonymous4trillion]

If you end up with an odd number when you add 1, find the average then add that on at the end

[@jazzboots8893]

I am never ambivalent about your videos! Always great!

[@whynotanyting]

"This video is sponsored by pie, tao, and e." hmmm pietaue? Oh, π τ e I'm going to bed

[@ADPuckey]

Jesse I'm so glad you're back. The E at the end made me lose my shit

[@Goodvvine]

I wish you can keep doing these videos, because they are really good

[@thror1709]

This is great content! glad i found your channel

[@CoryMck]

I figured this out in 3rd grade... _but I didn't get a story credited to my discovery_

[@defalt9405]

Im so glad that you came back to KZbin

[@Devynwithawhy]

I used this for a level up system in an rpg that I stoppex working on

[@owllover2319]

2:02 Counting is for Philistines!

[@sk8rdman]

I came across this same problem, but the way I solved it was that i imagined having a bunch of blocks, one for 1, two for 2, etc up to n. If I stack those blocks on top of each other, I can make them into a right triangle. Then to find out the total number of blocks, I just solve for the area of the triangle, which is half the area of an n*n square. The only nuance you have to worry about is that there's one extra row in the middle that adds an extra n/2 blocks. So my formula was (n^2+n)/2, which is algebraically the same as your formula. So for n=100 I just take (10000+100)/2=5050

[@JellyWaltzov]

this made my day. perfect explanation. Thank you!

[@kronlund420]

Great, as always

[@jotage3446]

I miss this channel being active :(

[@wepranaga]

when you see math as a problem solving rather than counting.

[@festivite]

Thank you my brain expanded from smol brain to big brain thank

[@splitscim]

Did anyone else come up with the equation before he told the story? Good practice for AP Comp Sci! Seems like an FRQ they'd put on the test

[@tverdyznaqs]

Huh, I didn't expect to laugh at a maths video! Good job dude, this was perfect

[@NateandNoahTryLife]

In 2019 I’m still dumber than a 7 year old from 1784... and that’s ok.

[@DozyBinsh]

We're all dumber than Gauss, tbh.

[@thesuperdoge2476]

Hey at least we're still alive

[@deidara_8598]

You have the world's knowledge at your fingertips, you can undo this whenever you like.

[@leocossham]

You learn how to do this in AS level maths in the UK

[@danielsteel5251]

This reminds me of when minutephysics did a FOIL video.

[@leocossham]

@@danielsteel5251 wait like foil as in what you wrap your sandwiches in or am I missing something

[@danielsteel5251]

@@leocossham en.wikipedia.org/wiki/FOIL_method

[@leocossham]

@@danielsteel5251 haha I understand the parallel there 😂😂 I remember this being taught as the smiley face method because you can sort of make the lines look like a smiley face?

[@WangleLine]

These videos are so good. I hope you'll return some day~

[@itisdevonly]

Match number pairs. 1 and 99, 2 and 98, 3 and 97, etc. That covers numbers 1-49 and 51-99 to make 49 number pairs that each add to 100, putting the total so far at 4900, then add in the 100 and 50 that were left out, and you get 5050. Did that calculation in a few seconds. Did I get it right? Maybe I don't need to watch past the 0:18 point if I came up with that already? Edit: Okay, finished watching the video. I wasn't far off. The solution he showed was a bit more general purpose than what I did, but it was pretty much the same idea.

[@DougStoddart]

great video - well done!

[@gangstabib]

Keep up the good work! This guy did it when he was a 7 year old kid!

[@platonic4ssploughing]

ok idk if im going crazy or not but i swear there was a video from this channel that had a section that critiqued a highschool textbook for its poor wording and credibility and had this giant ass wheel of fallacies where he shortly then made a gag about him never falling for these fallacies out of sheer superhuman perfection

[@connorking8503]

It works for odd-length sequences. All on it's own. Magic!

[@Vaaaaadim]

Indeed, it does happen to work for odd-length sequences, but even so it is worthwhile to show that this is true as well, because this doesn't always work out in general. Like for instance, there is a pattern that holds for the cosine power reduction formulas on the even powers that doesn't on the odd powers (but they are almost the same).

[@themasstermwahahahah]

Keep up the good work sir!

[@mb98765]

don't know how or when i subscribed, but i enjoyed the video

[@benjaminjackson8663]

Same

[@pabloaragon3303]

Same

[@TararidhKep]

3 more years until the next video.

[@Jaur-jaur]

OMG haven’t watched the video yet but I already love it!! Yay this channel is B-A-C-K!

[@misterkid]

I'm so glad I subscribed to you

[@nodisponible8]

love that endings

[@awsomebot1]

I already knew about this formula and it's story, but I've still watched this video multiple times. Please don't go again :^)

[@karthikkumar6861]

Hmm, I learned it slightly differently. Sum of any 'n' no of digits in a series is always (average X no. of digits). To find the average for a series you can do (First digit + Last digit)/2. So for this question Average would be (1+100)/2 = 50.5. Now sum would be 50.5 X 100 = 5050.

[@fr4ggle4]

New episodes of This Place?? Yes plz and thank you

[@blueicer101]

He’s right. I figured this when solving a maths problem in year 8. How many presents do you get from your true love in the 12 days of Christmas?

[@SiePhi]

But the even more generic form of the problem when the gap sizes vary is (LastNumber)=L and (FirstNumber)=F and (Gaps)=G is (L^2-F^2+L+F)/(2G)

[@Vaaaaadim]

And even more generic than that is to find the sum of a polynomial, and there are ways to come up with formulas for that as well.

[@SiePhi]

Where would one find this formula or a collection of them.

[@Vaaaaadim]

​@@SiePhi As it happens, this is a problem that I have had interest in for a very long time, and I had figured out multiple ways you can derive the formulas for this. There is a lot that I want to talk on this subject, and I intend to make a video about it one day. I'll show you what I think is the most straightforward way I've come up with. Note that I will only be talking about polynomials with integer powers. We are looking for a formula for the summation of a polynomial. Since a polynomial is a sum of powers of x multiplied by coefficients, we can split up the polynomial into its terms as separate summations and factor out the coefficients. So our problem is now essentially reduced to, can we find a formula for the summation from 1,...,n of x^p? If p = 1, we have an arithmetic sequence, p = 2 the perfect squares, so on so forth. The key observation is that, if we have some polynomial p(x), and we describe another function s(x) = the sum of the evaluations from 1,...,x of p(x), then it MUST be the case that s(x) - s(x-1) = p(x), and s(1) = p(1). Anyways, we can actually make a unique polynomial s(x) which can satisfy these conditions. Consider if I said s(x) = x^m, then s(x) - s(x-1) = x^m - (x-1)^m, and if you apply binomial expansion you will find that the largest powers cancel out and you get a polynomial of ONE LESS DEGREE, and furthermore, the highest powered term in this polynomial is m * x^(m-1). Cool, so lets use m = p + 1, and divide by p+1. But now we have all this other junk, x^(p+1) - (x-1)^(p+1) = (p+1)x^p + g(x). Why don't we just subtract a polynomial from x^(p+1) so that the g(x) part would be canceled out. In other words, subtract a polynomial which would be the summation of g(x) evaluated at 1,...,n. We can do this inductively and then we're set. I've typed up an example of going through these computations in LaTeX and have put them up as an imgur image imgur.com/a/H650vOy And here is those two examples that you can verify the validity of in desmos. www.desmos.com/calculator/lgzjtzdjh8 Let me know if further clarification is needed.

[@marsherr]

You are amazing

[@rudeus6621]

3:14 that sound effect tho😂

[@ksmyth999]

Do we know definitely that Gauss used the pairing method since in Wikipedia there seems to be some doubt? Also, his age is variously quoted as from 7-12. Many years ago I seem to remember reading that he was 10. Gauss was a child prodigy who could do complicated calculations in his head. This does not necessarily mean he was good at maths. But this story clearly demonstrated his promise, and he is now recognized as one of the greatest mathematicians of all time. There is another fairly simple method that he could have used. It is not quite so elegant but takes advantage of the symmetry in the numbers 1 to 100. I think he would have found this first and although it is a two-stage calculation if you include the thinking time, I believe he could have produced the correct answer just as quickly. It is fairly obvious, so I leave it as an exercise to the reader to work out.

[@samar5838]

Dude I learnt to do this in 10th grade.. its a simple arithmetic progression. You can even apply the formula n(n+1)/2 where n is the number of natural numbers taken starting from 1

[@hero19876]

Glad you aren't deceased

[@kidink0djillwill]

This works on even numbers to make pairs

[@blue_ouija]

"No, bitch." ~Carl Gauss

[@Matthew-hankla]

very nice

[@xexpo]

Would've been nice if you went slightly further with the sigma notation. For the 534->6389, you might've showed that you can do (1->6389) - (1->533), showing that the simplified n formula can be used for all.

[@TheReaverOfDarkness]

That would have been a third formula which is simpler than the first but more complex than the second.

[@jerry3790]

Tau should’ve been a full pie

[@Rovsau]

**Mathematicians vigorously refresh their ink pens**

[@jackthethinker]

Is this arithmetic progression?

[@pravinrao3669]

Well I figured it out. First I did (1+2+3+4+5+6+7+8+9)/9=5 I found out it if its odd and you do it its always middle term so I did it with 99 its 50. I multiplied 50 x99 = 4950 then I just added 100 so 5050

[@Camkitsune]

You pair up the numbers like so: 1 + 49, 2 + 48... Since there are 49 pairs you get 2,450 Add the 50 and 100 and you get 2,600.

[@itisdevonly]

Close, but not quite. You only paired up the numbers 1-49, which is not 49 pairs; it's only half that amount. You left out numbers 51-99 in your pair-adding system. You could instead pair 1 to 99, 2 to 98, etc. which actually does get you 49 pairs, but it's 49 pairs of 100. Add in the missing 100 and 50, and you get 5050. Alternatively, pair 1 with 100, 2 with 99, 3 with 98, etc. and you get 50 pairs of 101, which also gives you 5050.

[@LunchThyme]

You can do it the other way too, take the number in the middle and multiply by the total numbers. 50.5x100=5050

[@manuga2001]

Did you guys not learn arithmetic and geometric progression in highschool?

[@mort0303]

I have an A-level in mathematics and I was not taught this

[@manuga2001]

@@mort0303 seriously? For these progressions you only need to know about 6 formulas I think. I'm still in highschool and we learnt this not too long ago. I live in Australia btw.

[@Glorc72000]

Most high school math classes in the states and Canada are calculus and linear algebra based, where this isn't very relevant.

[@arnaldo8681]

We learn it in brazil as well

[@heliakopter]

@@Glorc72000 Wow, really? In my country we literally learn this in middle school and we also learn calculus and algebra in high school.

[@gecko2000405]

What sound does the equation make when the numbers work themselves out again?

[@Supernoxus]

I love your vids

[@Abhi-ib4fr]

Its basically an Arithmetic progression

[@mesplin3]

I like to call this the sandwich method. You take your list and fold it in half.

[@jjtt]

I remember when I was 15 that my then teacher showed us how to prove by complete induction that relationship for any maximum number. Then he proceeded to make us do 30 or 40 more similar proofs of that kind...

[@ArtArtisian]

Btw, i found the history of this story of gauss really interesting. Sorta a lense for how mathematical fables have changed over time. The original sum was probably shorter, the intent of the school teacher has been totally implied, and I'm pretty sure an early version mentions the risk of violence for students. See the following for a collection of versions, dated. bit-player.org/wp-content/extras/gaussfiles/gauss-snippets.html

[@nesrine7738]

Omg thank you so much, i finally get it✨✨✨😭🕊️

[@Junosensei]

I figured out the 1-100 sequence using the concept by framing it in my head (I didn't make a math sequence, though). I ignored the last 100 and did 1 + 99 = 100, 2 + 98 = 100, etc., which is 100 (49 times, because 100-1, then split in half, is 49.5), leaving 50 left. So... 4900 + the last 100 + the 50 left over = 5050. All while the video played. Finished before Carl did. But in high school, I took college algebra classes and stuff. We had a classmate who had the highest grade in our school and has since gone to work at NASA, and I still beat him in our state's big math test (as part of the regents), after having skipped the final few months of class and test prep due to studying abroad in Japan. The highest score in the school that year. After I left high school, I never took another math class again (except astronomy in college, where I decided to laze on a calculator). Get that freaking crap out of my life, plz. Ew. I love you, ThisPlace, but please stop. ;~;

[@robspiess]

Yeah, I basically did the same thing. 1-49 pairs up with 99-51 to make 49 sets of 100, so 4900, plus the 50 plus the 100 = 5050.

[@Vaaaaadim]

I would be very happy if he continued actually, people struggle with math and ThisPlace makes great explanations.

[@drthop]

Things change a little bit if the total of numbers is odd, but good video

[@murphygreen8484]

Wait. When did I subscribe to this channel? Guess I'll unsu......whaaa? Good video. Guess I'll stay subscribed!

[@snapfacts1525]

"No b$%#@" - Carl Gauss

[@Eyeofdust]

Simpler idea You know 1+100 is 101 And even when you get to the middle it’s 50+51=101 So let’s multiply 101 by 50 because we already reduced the numbers by half to add them together for all those 101s So in this case 50 being half of the latest number we had aka 100 And we get 5050 Or in 1+2+3+4 example 1+4 is 5 4/2 is 2 2*5=10

[@arjunpathak4582]

This is arithmetic progression, and i learnt this in school, its really easy im suprised he didnt know about it