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I've been having trouble finding the pattern and sum to a sequence that goes like this:

So when $c=3$, the sequence is $c(c-2)+(c-1)(c-1)+(c-2)c=2c(c-2)+(c-1)(c-1)$.

And when $c=4$, the sequence is $c(c-3)+(c-1)(c-2)+(c-2)(c-1)+(c-3)c=2(c-3)+2(c-1)(c-2)$

And when $c=5$, the sequence is $c(c-4)+(c-1)(c-3)+(c-2)(c-2)+(c-3)(c-1)+(c-4)c=2c(c-4)+2(c-1)(c-3)+(c-2)(c-2)$

So essentially there will be $c$ terms total. I substituted in the numbers for $c=3,4$ and got $3*1 + 2*2 + 1*3$ and $4*1 + 3*2 + 2*3 + 1*4$. This is not an arithmetic sequence or geometric sequence, but there is definitely a pattern. I want to know the pattern because I want to generalize it in order to find the sum of such a sequence where c equals some number.

Edit: Since this question might be confusing, I did find the solution for the summation of such a sequence, which is $c(c-1)(c-2)/6$. It was just that no context was provided for this answer. Even though it works for all c's that I plugged in, I have no idea how this summation formula came to be at all, and I was just looking for some kind of explanation for it.

The link to the website that I found this answer off of is this: https://cs.nyu.edu/mishra/COURSES/15.SOCNET/HW1s-15.pdf

It is for Q1 part b, which follows from the formula given in part a. It is a networks problem, but the part that I am having trouble understanding is not network-related at all but rather how they found the formula for the sequence. I managed to work out the second term in the numerator with the formula $n(a_1+a_n)/2$ (i.e., how one would find the sum of a sequence of integers going from 1 to 100), but I have no idea how they came up with the third term in the numerator.

Lucy
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  • It looks like you are expressing $\sum_{k=1}^n [k(n-k)].$ I don't know of any closed form expression for this, in terms of $n$; perhaps someone else does. I am confused, by your statement that (in effect) the summation would have $(n+1)$ terms; the summation only has $(n)$ terms. – user2661923 Feb 11 '21 at 05:57
  • Sorry, I made a mistake when I was looking at the sequence. You are right: the summation only has $n$ terms. – Lucy Feb 11 '21 at 05:58
  • $\sum_{k=1}^n k(n-k) = (n+1)n(n-1)/6 = \binom{n+1}{3}$ is well-known. It can be proven in a variety of ways, for instance using the first two cases of Faulhaber's formula, or induction, or the calculus of finite differences (thinking of it as a "discrete integral"), or generating functions and the binomial series, or by counting the number of points $(x, y, z)$ with $1 \leq x, y, z \leq n$ where $x+y+z \leq n+2$ in two ways using the "stars and bars" bijection. – Joshua P. Swanson Feb 11 '21 at 07:46
  • See https://math.stackexchange.com/questions/1113556/how-to-show-that-sum-k-1n-kn1-k-binomn23, and if you are interested, the OEIS sequence for Tetrahedral Numbers, https://oeis.org/A000292. – player3236 Feb 11 '21 at 07:46
  • @JoshuaP.Swanson While your math certainly pertains, my original mistake is infectious. Expression should be $$\sum_{k=1}^n k([n+1]-k).$$ – user2661923 Feb 11 '21 at 11:33

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