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December 9th, 2017, 04:56 PM  #1 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 614 Thanks: 83 Math Focus: Electrical Engineering Applications  Two Properties of Pascal's Triangle?
I was wondering about two properties of Pascal's Triangle (PT) that appear to hold true (as far as I have checked), and if they are indeed known properties. The properties involve power sums to 0 and power sums to the factorial of a row number, using the values of PT as coefficients. Unfortunately, Googling with "Pascal's Triangle", "sum", and "factorial" leads to a lot of noise to try to filter through, due to the nature of the terms of PT. So I thought I would ping the forum to check for any existing knowledge. The reason that I am asking is related to this thread as I will explain in another thread that I will start involving the solution to these types of problems. (I will also post another possible way of obtaining the solution in the referenced thread). For reference, below is a portion of the left justified PT: $\displaystyle \large {\begin{array}{c c} & k & 1 & 2 & 3 & 4 & 5 & 6 \\ r \\ 0 & & 1 \\ 1 & & 1 & 1 \\ 2 & & 1 & 2 & 1 \\ 3 & & 1 & 3 & 3 & 1 \\ 4 & & 1 & 4 & 6 & 4 & 1 \\ 5 & & 1 & 5 & 10 & 10 & 5 & 1 \end{array}}$ The first property is: For a given row $r$ and for each $k^{th}$ term in $r$ as given above, for any number $n$ (although I am only interested in $n \geq r$): $\displaystyle \large PT_{r,1} \cdot n^r  PT_{r,2} \cdot (n1)^r + PT_{r,3} \cdot (n2)^r ... PT_{r,r+1} \cdot (nr)^r = r!$ (alternating signs for each term). For example, with $n=5$ and $r=4$: $\displaystyle \large 1 \cdot 5^4  4 \cdot 4^4 + 6 \cdot 3^4  4 \cdot 2^4 + 1 \cdot 1^4 = 4! = 24 $ and with $n=6$ and $r=4$: $\displaystyle \large 1 \cdot 6^4  4 \cdot 5^4 + 6 \cdot 4^4  4 \cdot 3^4 + 1 \cdot 2^4 = 4! = 24 $ so: $\displaystyle \large 1 \cdot n^4  4 \cdot (n1)^4 + 6 \cdot (n2)^4  4 \cdot (n3)^4 + 1 \cdot (n4)^4 = 4! = 24 $ which can be shown to be true after expanding and simplifying. This property is true for all of the values that I have checked numerically so far (up to $r=100$ for $n=r+1$ and for various other values for $r,n$). I can see why it might be true since PT is instrumental in calculating $(na)^r$, but I really don't have the skills to prove it analytically. The second property is: For a given row $r \gt 0$ and for each $k^{th}$ term in $r$ as given above, for any number $n$, although I am only interested in $n \geq r$ (and $rq \geq 2$): $\displaystyle \large PT_{r,1} \cdot n^{(rq)}  PT_{r,2} \cdot (n1)^{(rq)} + PT_{r,3} \cdot (n2)^{(rq)} ... PT_{r,r+1} \cdot (nr)^{(rq)} = 0, \quad 1 \leq q \leq r$ (alternating signs for each term). For example, with $n=5$, $r=4$, $q=1$: $\displaystyle \large 1 \cdot 5^3  4 \cdot 4^3 + 6 \cdot 3^3  4 \cdot 2^3 + 1 \cdot 1^3 = 0$ and with $n=5$, $r=4$, $q=2$: $\displaystyle \large 1 \cdot 5^2  4 \cdot 4^2 + 6 \cdot 3^2  4 \cdot 2^2 + 1 \cdot 1^2 = 0$ and with $n=7$, $r=5$, $q=3$: $\displaystyle \large 1 \cdot 7^2  5 \cdot 6^2 + 10 \cdot 5^2  10 \cdot 4^2 + 5 \cdot 3^2  1 \cdot 2^2 = 0$ Again, I do not possess the skills to prove/disprove for all $n, r, q$. Obviously, comments are welcome. 
December 9th, 2017, 09:09 PM  #2 
Senior Member Joined: May 2016 From: USA Posts: 916 Thanks: 366 
I am rather tired and am having some difficulty understanding your notation. Using standard notation, the kth term of the nth row is $\dbinom{n}{k} \equiv \dfrac{n!}{k! * (n  k)!} \text {, where } 0! = 1 \text { and } m! = m * (m  1)! \text { if } m \in \mathbb Z \text { and } m > 0.$ 
December 9th, 2017, 09:43 PM  #3 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 614 Thanks: 83 Math Focus: Electrical Engineering Applications 
Oops, I was so caught up in my application with terms starting at 1 that I did not follow convention by starting $k$ at $0$. Plus, I used $r$ for $n$ (but at least it starts at 0). I will have to ask readers to bear with my mistakes. My application is hopefully explained here. Last edited by jks; December 9th, 2017 at 09:58 PM. Reason: Add reference to the problem I am trying to solve. 

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