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December 9th, 2017, 05:56 PM  #1 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 634 Thanks: 96 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, 10:09 PM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,300 Thanks: 549 
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, 10:43 PM  #3 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 634 Thanks: 96 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 10:58 PM. Reason: Add reference to the problem I am trying to solve. 
June 17th, 2018, 01:28 PM  #4 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 634 Thanks: 96 Math Focus: Electrical Engineering Applications 
Both of these appear to be true due to Euler's Finite Difference Theorem as can be seen here in Equation 10.1 on page 45. As stated in the reference: $\displaystyle \sum_{k=0}^n (1)^k \binom{n}{k} k^j = \begin{cases} 0, & 0 \leq j < n \\ (1)^n n!, & j=n \end{cases} $ Of course, we can make the factorials all positive (largest $k$ is positive) by multiplying each side by $(1)^n$: $\displaystyle \sum_{k=0}^n (1)^{n+k} \binom{n}{k} k^j = \begin{cases} 0, & 0 \leq j < n \\ (1)^{2n} n!, & j=n \end{cases} $ Since $(1)^{2n}=1$, we get: $\displaystyle \sum_{k=0}^n (1)^{n+k} \binom{n}{k} k^j = \begin{cases} 0, & 0 \leq j < n \\ n!, & j=n \end{cases} $ and multiplying the left side by $(1)^{2k}=1$, for each $k$ in turn, we get: $\displaystyle \sum_{k=0}^n (1)^{nk} \binom{n}{k} k^j = \begin{cases} 0, & 0 \leq j < n \\ n!, & j=n \end{cases} $ which is the form that I have recently been working with. 

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