I was learning the derivatives and I remembered Pascal's triangle and I discover that I can write it with derivatives without having to draw it.
Let's see Pascal's triangle:
Any row can be written using derivatives, per example the $6th$ row.
The algorithm is like this: You start with $x^n$ where $n =$ Number of row.
And every time you calculate the derivative, you divide that same derivative into the number of times you've derived.
I'll write the $5th$ row:
$x^5$,
1° derivative: $\frac {d}{dx}x^5 = \frac{5x^4}{1}$ ( I Divide in $1$ because is the first derivative) Here i get the first term of Pascal's triangle.
2° derivative: $\frac {d}{dx}5x^4 = \frac{20x^3}{2} = 10x^3$ Here i get the second term of Pascal's triangle.
3° derivative: $\frac {d}{dx}10x^3 = \frac{30x^2}{3} = 10x^2$ Here i get the third term of Pascal's triangle.
4° derivative: $\frac {d}{dx}10x^2 = \frac{20x}{4} = 5x$ Here i get the four term of Pascal's triangle.
In fact, writing algebraically and with a little more work I have come to another formula more complex:
$A_k = \frac{P^{(k-1)* n! * (rx)^{(n-k+1)}}}{(n-k+1)! * (k - 1) !}$
With which I can get any term from a binomial of the form $(rx + p)^n$
Although, I do not think it works much, but I found it interesting.
And I would like to know why this happens, thank you very much.
