Evaluate the sum using derivatives and generating functions Evaluate $\sum_{k=1}^{n-3}\frac{(k+3)!}{(k-1)!}$. 
My strategy is defining a generating function,
$$g(x) = \frac{1}{1-x} = 1 + x + x^2...$$
then shifting it so that we get,
$$f(x)=x^4g(x) = \frac{x^4}{1-x}= x^4+x^5+...$$
and then taking the 4th derivative of f(x).
Calculating the fourth derivative is going to be a little tedious but it won't be as bad compared to the partial fraction decomposition I will end up doing. What is a better way to evaluate the sum using generating functions?
 A: If you consider $U_4(k)= k(k+1)(k+2)(k+3)(k+4)$
Then $U_4(k)-U_4(k-1)=k(k+1)(k+2)(k+3)\bigg((k+4)-(k-1)\bigg)=5U_3(k)$
Thus you get a telescoping series and
$$\sum\limits_{k=1}^{n-3}k(k+1)(k+2)(k+3)=\dfrac 15\bigg(U_4(n-3)-U_4(0)\bigg)=\dfrac 15U_4(n-3)\\=\dfrac{(n-3)(n-2)(n-1)(n)(n+1)}5$$
A: When you differentiate $g(x)$ $4$ times all powers of $x$ initially less than $4$ disappear anyway. 
Then:
$$g^{(4)}(x)=4!(1-x)^{-5}=\sum_{k\ge 0}\frac{(k+4)!}{k!}x^k$$
Operating with $(1-x)^{-1}=1+x+x^2+x^3+\cdots$ on both sides gives a new expansion with terms which are partial sums of the coefficients of $x^k$ in the previous expansion:
$$(1-x)^{-1}g^{(4)}(x)=\sum_{r\ge 0}\left(\sum_{k=0}^{r}\frac{(k+4)!}{k!}\right)x^r$$
so
$$4!(1-x)^{-6}=\sum_{r\ge 0}\left(\sum_{k=0}^{r}\frac{(k+4)!}{k!}\right)x^r$$
$$\implies [x^r]4!(1-x)^{-6}=4!\binom{r+5}{5}=\sum_{k=0}^{r}\frac{(k+4)!}{k!}$$
but $r=n-4$ to match up with $n$ in the question, so

$$\sum_{k=0}^{n-4}\frac{(k+4)!}{k!}=4!\binom{n+1}{5}\tag{Answer}$$

This summation is the same as the one in the question with only a shift in summation index.
Of course you may notice that this is just Pascal's hockey stick rule.
A: Here is a variation using generating functions without differentiation. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. This way we can write for instance
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}\tag{1}
\end{align*}

We obtain for $n\geq 4$
  \begin{align*}
\color{blue}{\sum_{k=1}^{n-3}\frac{(k+3)!}{(k-1)!}}
&=4!\sum_{k=1}^{n-3}\binom{k+3}{4}=4!\sum_{k=0}^{n-4}\binom{k+4}{4}\tag{2}\\
&=4!\sum_{k=0}^{n-4}[x^4](1+x)^{k+4}\tag{3}\\
&=4![x^4](1+x)^4\sum_{k=0}^{n-4}(1+x)^k\tag{4}\\
&=4![x^4](1+x)^4\frac{(1+x)^{n-3}-1}{(1+x)-1}\tag{5}\\
&=4![x^5](1+x)^{n+1}\tag{6}\\
&\,\,\color{blue}{=4!\binom{n+1}{5}}\tag{7}
\end{align*}

Comment:


*

*In (2) we write the fraction using binomial coefficients and shift the index by one to start with $k=0$.

*In (3) we use the coefficient of operator according to (1).

*In (4) we use the linearity of the coefficient of operator.

*In (5) we apply the finite geometric series formula.

*In (6) we do some simplifications, apply the rule $[x^{p+q}]A(x)=[x^p]x^{-q}A(x)$ and ignore $(1+x)^{4}$ since it does not contribute to $[x^5]$.

*In (7) we select the coefficient of $[x^5]$ accordingly.
