I have the following expression:


Without the $i!$, the above expression would simply reduce to $(2x)^n$, but is there a way, or method for simplifying the expression which includes the extra $i!$? I mean, I could simplify to $$\sum_{i=0}^{n}\frac{n!}{(n-i)!}(2x+1)^{n-i}(-1)^i$$ But this is not really simplifying as I am trying to find a way to remove the sigma.

  • $\begingroup$ The question is not clear... $\endgroup$ – k1.M May 14 '15 at 13:44
  • $\begingroup$ Is there a way to simplify the expression which includes the $i$? $\endgroup$ – Iceman May 14 '15 at 13:51
  • $\begingroup$ I guess that the first expression is simple enough, and more elegant...so you have a closed form for a summation, what's the problem? What you need? Whats the reason for simpler expression? $\endgroup$ – k1.M May 14 '15 at 13:55
  • $\begingroup$ I know, but without the $i!$ it becomes $(2x)^n$ which is much simpler... My problem involves three nested sums and I'm just trying to reduce (which I'm not sure is possible) as best as I can. $\endgroup$ – Iceman May 14 '15 at 13:57

By setting $j=n-i$, $$\sum_{i=0}^{n}\frac{n!}{(n-i)!}(2x+1)^{n-i}(-1)^i = (-1)^n n!\sum_{j=0}^{n}\frac{(-1)^j (2x+1)^j}{j!},$$ hence your sum is just $(-1)^n n!$ times a partial sum for the Taylor series of $e^{-(2x+1)}$ in a neighbourhood of the origin.


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