# Binomial Theorem…extra indexed term.

I have the following expression:

$$\sum_{i=0}^{n}\binom{n}{i}(2x+1)^{n-i}(-1)^ii!$$

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.

• The question is not clear... – k1.M May 14 '15 at 13:44
• Is there a way to simplify the expression which includes the $i$? – Iceman May 14 '15 at 13:51
• 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? – k1.M May 14 '15 at 13:55
• 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. – 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.