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Given $p$,$x$ and $n$ , I have to calculate(find a closed form for the following four series) :

1) $p^1+p^2+p^3+.......+p^n$

2) $p^1(p^1-1) +p^2(p^2-1)+p^3(p^3-1)+.......+p^n(p^n-1)$

3) $p^1(p^1-1)(p^1-2)(p^1-3) +p^2(p^2-1)(p^2-2)(p^2-3)+p^3(p^3-1)(p^3-2)(p^3-3)+.......+p^n(p^n-1)(p^n-2)(p^n-3)$

4) $p^1(p^1-1)(p^1-2)(p^1-3)....(p^1-x) +p^2(p^2-1)(p^2-2)(p^2-3)........(p^2-x)...+p^3(p^3-1)(p^3-2)(p^3-3)......(p^3-x)+............+p^n(p^n-1)(p^n-2)(p^n-3)........(p^n-x)$

I was able to calculate the first two series using sum of G.P. method. Looking for the solution of last 2 series.

Sanghty reqwwe
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1 Answers1

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For the third one, just multiply out.

$$\begin{align*}\sum_{k=1}^n p^k(p^k-1)(p^k-2)(p^k-3) & = \sum_{k=1}^n\left(p^{4k}-6p^{3k}+11p^{2k}-6p^k\right) \\ & = \sum_{k=1}^n(p^4)^k-6\sum_{k=1}^n(p^3)^k+11\sum_{k=1}^n(p^2)^k-6\sum_{k=1}^np^k\end{align*}$$

Now, each is a geometric sum.

SlipEternal
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  • For the last one, I don't see how you would have a closed form for that. It appears to use Stirling numbers of the first kind, but I don't see any simplification from that. – SlipEternal Dec 10 '19 at 18:42
  • For, x=3 , the coefficients of the final answer are :- [1,-6,11,-6] as that can be seen from your answer , I want to know if there is some efficient way to calculate coefficients set for , say, x=100 . – Sanghty reqwwe Dec 10 '19 at 18:46
  • Stirling Numbers of the First Kind. For $x=100$, here are the coefficients: Wolframalpha – SlipEternal Dec 10 '19 at 19:30
  • And in general, to get a table of the coefficients from Wolframalpha, it would be: $$\text{Table[StirlingS1[x+1, (x+1-k)],{k,0,x}]}$$ – SlipEternal Dec 10 '19 at 19:36