why the fuck does pascal's triangle correlate with the coefficient

>>8802851
because it is literally the same thing

>>8802853
thanks

>>8802851
the rows are calculated using the same operation you could use to calculate binomial coefficients.

Imagine finding (1+x)^k by expanding (1+x)^(k-1) and then multiplying it by 1+x, and collecting like terms.

When you collect like terms you're adding two consecutive coefficients from the expansion of (1+x)^(k-1).

Just like when you find the kth row of Pascal's triangle you add two consecutive entries of the k-1th row

>>8802874
To add to this, you can create a generating function from the sum, giving you the fibonacci sequence

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(a + b ) ^ 3= aaa + 3 * (aab) + 3 * ( abb) + bbb

I have 3 positions, xxx, and I can assign each one to a or b. How many ways can I do this.

aaa - 1
aab, aba, baa - 3
bba, bab, abb - 3
bbb -1

>>8802925
this explains why the binomial coefficients are the same as n choose k. It doesn't really explain why Pascal's triangle works.

>>8802980
Pascal's triangle is effectively a representation of the combination function

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>>8802851
>why
Deus vult

>>8803061
Deus Vult is a normalfag meem killyourslf

>>8803070
I'm sorry I've offended your leftist principles

I clearly meant inshallah

>>8803082
You don't even know how to use it. Fuck off zlumpflet

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>>8803087
>zlumpflet
???

Using the notation
[eqn] {n\choose k} = \frac{n!}{k! (n-k!)}[/eqn]
We will assume the binomial expansion is true
[eqn] (x+y)^n = \sum_{k=0}^n {n \choose k}x^k y^{n-k} [/eqn]
Multiplying both sides by [math] x+y [/math] gives
[eqn] (x+y)^{n+1} \ \ = \sum_{k=0}^{n} {n\choose k}x^{k+1} \ \ y^{n-k}\ \ + \sum_{k=0}^n {n \choose k} x^k y^{n+1-k} \\ = \sum_{k=1}^{n+1} {n\choose k-1}x^k y^{n+1-k} \ \ \ + \sum_{k=0}^n {n\choose k}x^k y^{n+1-k} [/eqn]
If for convenience use the definition that [math] a!=\infty [/math] if [math] a [/math] is a negative integer we can combine these to get
[eqn] (x+y)^{n+1} \ \ = \sum_{k=0}^{n+1} \Bigg({n \choose k -1} + {n \choose k}\Bigg)x^k y^{n+1-k} \ \ \ = \sum_{k=0}^{n+1} {n+1 \choose k}x^k y^{n+1-k}[/eqn]
where the far RHS is true by assumption as it's simply the binomial expansion of [math] (x+y)^{n+1} [/math]
Now, here's a very useful mathematical theorem anon. If you have two polynomials, finite or infinite, the only way they are equal is if all the respective coefficients for each term are equal. This means

[eqn] {n \choose k-1} + {n \choose k} = {n +1 \choose k} [/eqn]
which is nothing more than Pascal's triangle.

>>8802851
You can prove that [math]\binom{n}{a} + \binom{n}{a+1} = \binom{n+1}{a+1}[/math] but I can't be bothered to so do your own homework.
From that, and a generalized binomial expansion you can demonstrated that property