New number triangle? I was dealing with taking derivatives from the form of $\sum_{i=1}^n x*(1-x)^i$ , where $n$ goes to infinity. When $n$ gets larger, you quickly arrive to very long functions with big and many coefficients, definitely in the derivative. So I tried to look for a pattern in it. I managed to find a pattern and in the core of this pattern, I found a number triangle.
I found the pattern to first group all the coefficients that had the same power of $x$. In this number sequence I found out that taking the derivative $i+1$ times, you could easily, through simple addition, find all the coefficients as $n$ goes up. Obviously, I would therefore, as $n$ gets larger, first need to find the first number of coefficients. And if $n$ goes up to infinity, you still needed an infinite amount of time to do it. And now it gets difficult to explain. If you could find the first $i+1$ derivatives, you could work backwards and complete the number sequence of the coefficients to infinity, without having to calculate the first number of derivatives of the original function. And that's when I made a number triangle and I could see a pattern in it. These are the first rows of the triangle;

And then you could see that you can make the triangle by starting at 1 and go to left (or right) and keep adding 1. When that is done, you can perform the following calculation: $\frac{(row*row_{-1})}{(row-row_{-1})}$. An example:
For the 7th row: you start with 7. Then you do $\frac{6*7}{7-6} = 42$ then you get $\frac{42*30}{42-30} = 105$ and so on. Using this triangle you can easily find the coefficients of the derivative of the function: $\sum_{i=1}^n x*(1-x)^i$ 
For example: suppose you complete take the 4th row: $ 4,12,12,4$
This is what you do next:
$4+12= 16$, $ 12+12+16 = 40$ , $ 12+4+24+40 = 80$. Now, the last number $'4'$ keeps repeating itself in the end, so the next step is $4+16+40+80 = 140$. and so on.
As you will see, if you take the derivative of $\sum_{i=1}^n x*(1-x)^i$ with for example $n = 7$, the number sequence of the coefficients with x to the power of $i= 3$ 
are $4,16,40,80,140$. 
What I found was that this is related to Pascal's triangle, where you just multiply the rows of Pascal's triangle times the number of the row!
Now, some questions to you all:


*

*Is there another way that already has been 'invented' to find the derivative that kind of function in another, easy way?

*Is this 'just' something new about the Pascal's triangle that we didn't knew yet?

*Is this useful??
I'm pleased to get feedback on this. Also, if the explanation is rubbish, I'll edit it, but you do can give feedback in how to explain it here. 
thanks
 A: What you have is precisely Pascal's triangle, but with the $n^{\text{th}}$ row multiplied by $n+1$ for each $n \ge 0$; that is, the $k^{\text{th}}$ entry in row $n$ is $(n+1) \dbinom{n}{k}$.
To see this, note that
$$
\begin{align*}
& \frac{d}{dx} \sum_{i=0}^n x(1-x)^i \\
&= \frac{d}{dx} \sum_{i=0}^n \sum_{r=0}^i (-1)^r \dbinom{i}{r} x^{r+1} && \text{binomial theorem} \\
&= \sum_{i=0}^n \sum_{r=0}^i (-1)^r(r+1)\dbinom{i}{r} x^r && \text{differentiating}
\end{align*}
$$
For fixed $k$, the coefficient of $x^k$ is therefore given by:
$$
\begin{align*}
&\sum_{i=0}^n (-1)^k (k+1) \dbinom{i}{k} \\
&= (-1)^k (k+1) \sum_{i=0}^n \binom{i}{k} && \text{pulling out constant factors} \\
&= (-1)^k (k+1) \binom{n+1}{k+1} && \text{by a combinatorial identity} \\
&= (-1)^k (n+1) \binom{n}{k} && \text{by a combinatorial identity}
\end{align*}
$$
So you see that the $k^{\text{th}}$ entry in the $n^{\text{th}}$ row of your triangle is (if you ignore the sign) just $n+1$ times the $k^{\text{th}}$ entry in the $n^{\text{th}}$ row of Pascal's triangle.
Is it new? Probably not. Identities like this arise all over the place in combinatorics (see generating functions and combinatorial species).
Is it useful? Well kind of, but on its own, not really; it's useful for computing the derivative of $\displaystyle \sum_{i=0}^n x(1-x)^i$, though.
