Why is the sum of numbers equal to N choose 2?

There are many ways to prove the sum of numbers from 1 to n, but this one is especially interesting. #math #maths #mathematics #shorts

Math overflow
Wolfram Demonstrations project
Larson, Loren C. "A Discrete Look at 1 + 2 + ⋯ + n."

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[MindYourDecisions]

Thanks for sharing the alternate proofs, like pairing numbers in twice the sum. I think most people learn a proof by induction in school as well. I was always amazed at the coincidence that the formula is equal to (n + 1) choose 2. This visualization shows the combinatorial interpretation. It's not simple at all, which is why it's remarkable someone thought of it. Credit to the paper Larson, Loren C. "A Discrete Look at 1 + 2 + ⋯ + n."

[k_Sub_challenge_with_no_video]

Idk why it looks more confusing than numerical derivation of that formula

[tamirerez]

Gauss did it differently, much simpler, much more visual.
Suppose: for the sum of the numbers from 1 to 6.
He completed the triangle into a rectangle. By adding another inverted triangle of 6 to 1
A rectangle with a length of 7 and a width of 6 was obtained. nx(n+1)
The area of ​​the rectangle is 42.
which is actually the area of ​​the two triangles.
Therefore 42:2 and we got 21

[phalanxutsav]

Nice visualization
I do it by another method -
Suppose our series is starting from 1 and ending at n then the sum of first and last term is n+1 and the sum of second and second last term is also n+1 and so on and it is for n/2 times so -> n(n+1)/2

[efegecili]

I keep it in my mind by: "Arithmetical mean×number of terms".

[rqukwwc]

I made a upside-down triangle next to the triangle.
Then, there will be a parallelogram there and the base is n+1.
The area of parallelogram is base times hight.
In this parallelogram, it will be (n+1)×n
The area of triangle is the half of it
so it is n(n+1)/2
I'm japanese 9th grader and I'm not good at English.

[fahad_hassan_]

I think of it as:
Flip the triangle vertically and then add those two triangles together. It is a well known fact that a rectangle or square can be made by join two triangles, if use this fact in reverse, we can imagine the two triangles as a rectangle, with length = n + 1 (as the max numbers of circles is n + 1)
and width = n (number of rows)
So area = n(n+1) and since the area of triangle is half of the rectangle, area = n(n+1)/2

However the merging of two triangles part lacks integrity even if it works

[spacemanspiff]

Every time I want to derive this fact, I just make an nxn "staircase" of squares and find the area.

[finnwilde]

another may of visualising it would to be to add another one of those beige triangles of the sum 1 - n, adjacent to the first to form a parallelogram of sides n and n+1. Then we work out area of parallelogram as n(n+1) and divide by 2 after to get sum of 1-n --> n(n+1)/2

[VSN]

This proof is an excellent introduction to the bijective principal

[Tymon]

I'm sorry Presh, I normally love your videos and explanations, but this example is really not the easiest to understand

[Agent_Ax]

I proved this for my self using Arithmetic Progression

Let us assume an AP
1, 2, 3, 4...n

So we can see that a=1, d=1
Where a and d are the first term and common difference

We know that sum of an AP is
Sn=(n/2)[2a+(n-1)d]
Sn=(n/2)[2*1+(n-1)*1]
Sn=(n/2)[2+n-1]
Sn=(n/2)[n+1]
Sn=n(n+1)/2

Hence proved

[igorfujs]

Much simpler is to visualize to add first and last row, then second and second last and so on. So for n numbers the pair's sum would be n+1. And you have n/2 pairs. So the result is (n+1)*n/2

[psibarpsi]

You can derive it using recursive definition of the sum also. That's also elegant to a degree.

[yuanyuanshen]

Another visual proof:
Add another identical triangle, flip it upside down, and put it on the side of the original triangle.
Now we have n rows, and each row has n+1 circles. The total is n(n+1).
So the original triangle is half of the total: n(n+1)/2.

[MmKayUltra]

Dude I love this I'm a bit of a pascals triangle buff and one of my favorite bits of it is its self referential third column. This might get me to be able to show people why I like it so much. It also is a great way to show how this extends into higher dimensions

[carloscrespo-riera]

I Iike the Quod Erat Demonstrandum at the end.

[abhijitray]

This channel is amazingly being brief enow to evaluate out any difficult calculations in a very basic way... And that to be by means of visualisation of different shapes with colourful exact structures geometrically, forming a simpler generalised standard algebraic equation.. Tons of Love nd Support to this Channel as well as the speaker/host Presh..

[MangoLassiYT]

there is easier proof of grouping 1st and last element and the total groups are n/2 and sum of 1st and last element is n+1
then total sum is n/2.(n+1) cause sum of every group is (n+1)
LETS MAKE IT EASIER WITH EG
n = 6
1 2 3 4 5 6
1+6=2+5=3+4=7
so pairing last and first elements gives same sum
also note there are n/2 groups and sum of each group is (n+1)so we can form total sum using n/2 *(n+1)
For odd n
n=5
1 2 3 4 5
here 3 remains ungrouped
if we do n/2 we get 2.5
there are 2.5 groups so where did 0.5 came from
remembee half of a group is same as remaining element
to check that (5+1)/2=3 which is half group and third element is also 3 so the third elemtn is the half group
this way proof is completed.if you didn't understand i tried my best

[blizzsoft]

When i see this at the first time i didn't understand. but at last i have understood this and it looks amazing