Coefficient of $x^{50}$ in the expansion of $\prod_{n=1}^{52}{(x+n)}$ Find the coefficient of $x^{50}$ in the expansion of 
$$\prod_{n=1}^{52}{(x+n)}$$
I can't find a way out. 
 A: Let $f(x)=\Pi_{n=1}^{52}(x+n)$. Then $f(x)$ is a monotic polynomial with degree 52 and roots $x_1=-1,x_2=-2,x_3=-3,...,x_{52}=-52$. By Vieta's formula, the coeficient of $x^{50}$ is 
$$x_1x_2+x_1x_3+...+x_1x_{52}+x_2x_3+x_2x_4+...+x_2x_{52}+...+x_{51}x_{52}$$.
Thus:
$\begin{eqnarray}
&&x_1x_2+x_1x_3+...+x_1x_{52}+x_2x_3+x_2x_4+...+x_2x_{52}+...+x_{51}x_{52}\\
&=&x_1(x_2+x_3+...+x_{52})\\
&&+x_2(x_3+x_3+...+x_{52})\\
&&+x_3(x_4+x_5+...+x_{52})\\
&&\vdots\\
&&+x_{50}(x_{51}+x_{52})\\
&&+x_{51}x_{52}\\
&=&x_1(x_1+x_2+x_3+...+x_{52}-x_1)\\
&&+x_2(x_1+x_2+x_3+...+x_{52}-x_1-x_2)\\
&&+x_3(x_1+x_2+x_3+x_4+x_5+...+x_{52}-x_1-x_2-x_3)\\
&&\vdots\\
&&+x_{50}(x_1+x_2+x_3+x_4+x_5+...+x_{52}-(x_1+x_2+x_3+x_4+x_5+...+x_{50}))\\
&&+x_{51}(x_1+x_2+x_3+x_4+x_5+...+x_{52}-(x_1+x_2+x_3+x_4+x_5+...+x_{51}))
\end{eqnarray}$
Now it is easy to conclude
A: As the coefficient of $x^{j}$ in a polynomial $P(x)$ is ${1\over j!}{d^j\over dx^j}P(x)|_{x=0}$,
coeff of $x^{50}$ in $\prod_{n=1}^{52}(x+n)$ 
= coeff of $x^{2}$ in $\prod_{n=1}^{52}(1+nx)$
=${1\over2!}{d^2\over dx^2}(\prod_{n=1}^{52}(1+nx))\Big{|}_{x=0}$
$$
{1\over2!}{d^2\over dx^2}\prod_{n=1}^{52}(1+nx)\Big{|}_{x=0}={1\over2}{d\over dx}\left[\sum_{n=1}^{52}n\prod_{j\neq n}(1+jx)\right]\Big{|}_{x=0}\\
={1\over2}\left[\sum_{n=1}^{52}n\sum_{j\neq n}j\prod_{k\neq n,j}(1+kx)\right]\Big{|}_{x=0}={1\over2}\left[\sum_{n=1}^{52}n\sum_{j\neq n}j\right]\\
={1\over2}\left[\sum_{n=1}^{52}n\left({52\times53\over2}-n\right)\right]
={1\over2}\left[\left({52\times53\over2}\right)^2-{52\times53\times105\over6}\right]\\=925327
$$
A: If $n$ is the number of terms being multiplied together ($52$ in your question) and $k$ is the target exponent ($50$ in your question), then the corresponding coefficient is $[{n+1 \atop k+1}]$ where $[{n \atop k}]$ is a Stirling number of the first kind. Note that the algebraic proof contains the same question you are asking.
A: From a more concrete perspective, you can only form a $x^{50}$-term by choosing exactly two factors from which to extract constant values, that is, you choose two parentheses $(x+n)$ from which to pick the $n$:s, and pick $x$:s from the rest. So the coefficient of $x^{50}$ is the sum of all products $ab$ where $a\neq b$ and $a,b\in \{1,2,\ldots,52\}$ (with no double-counting, i.e. $a<b$). A tedious computation (or a small computer program :) ) would conclude this with the answer.
Edit -  Here's the computation in question:
$$\sum_{a=1}^{52}a\sum_{b=a+1}^{52}b=\sum_{a=1}^{52}a\frac{(52+a+1)(52-a)}{2}$$
$$=\sum_{a=1}^{52}a\frac{52^2-a^2+52-a}{2}=\frac{1}{2}\sum_{a=1}^{52}(52^2a-a^3+52a-a^2)$$
$$=\frac{1}{2}\Big( (52^2+52)\frac{52(52+1)}{2} -\Big(\frac{52(52+1)}{2}\Big)^2-\frac{52(52+1)(2\cdot 52+1)}{6} \Big)$$
$$=...(arithmetic)...=925327$$
A: Let $S=\{1,2,\ldots,52\}$. And let $c_n$ denote the coefficient of $x^n$ in our polynomial's expansion. (for convenience)
We then have $\displaystyle \prod_{r \in S}(x+r)=\sum_{n=0}^{|S|=52}e_{52-n}(S)x^{n} $ where $e_n(r_1,r_2,\ldots,r_{52})$ is the $n$-th elementary symmetric polynomial on $52$ variables. Hence we have $\displaystyle c_{50}=e_2(S)=\sum_{\substack{i,j\in S \\  i<j}} i j$.
Then, we do some algebraic manipulations on $ \left(\sum S \right)^2$ to get

$$\begin{align} \left( \sum_{n \in S} n \right)^2 &= \sum_{i,j \in S}i
 j \\  &= \underbrace{\sum_{(i<j)\in S}ij+\sum_{(i>j)\in S}ij}_{2
 e_2(S)} +\underbrace{\sum_{(i=j)\in S}ij}_{\sum n^2} \end{align}$$

And finally

$$\begin{align} c_{50} &= \frac{1}{2} \left(
 \left(\sum_{n=1}^{52}n\right)^2-\sum_{n=1}^{52}n^2\right) \\   &={52^2
 \cdot(52+1)^2 \over 2 \cdot 2^2}-{52 (1+52) (1+2 \cdot 52) \over 2
 \cdot 6}\\  &= 925327 \end{align}$$


Addendum: Faulhaber's formula and the fundamental theorem  of symmetric polynomials guarantee that this same reasoning/algorithm works for any coefficient $c_n=e_{d-n}(1,2,\ldots,d)$ of $\prod_{i=1}^d (x+i)$. However, the complexity rises rapidly as the degree of the symmetric polynomial rises. Already for $e_5(S)=c_{47}$, we have this quagmire, where $\varsigma_n$ will stand for $\sum_{i=1}^{52} i^n$

$\begin{align} 
 e_5(S) & =  {1 \over 120} \left(  \varsigma _1^5-10 \varsigma _2 \varsigma _1^3+20 \varsigma _3 \varsigma _1^2+15 \varsigma _2^2 \varsigma _1-30 \varsigma _4 \varsigma _1-20 \varsigma _2   \varsigma _3+24 \varsigma _5\right)  \\
 & = 31848946440220
 \end{align}$

(for a more systematic way of calculating such results, use Newton's identities recursively)
Fortunately though, this family of polynomials and their coefficients are very well studied. These polynomials $x^{(n)}$ are called the Pochhammer Polynomials, and we have $\displaystyle x^{(n)}=\prod_{k=0}^n (x+k)=\sum_{k=0}^n \left[ {n \atop k} \right]$, where $\left[ {n \atop k} \right]$ are the unsigned Stirling numbers of the first kind. Stirling numbers are relatively speaking much easier to calculate using the recursive relationship $\left[ {n+1 \atop k} \right]=n \left[ {n \atop k} \right]+\left[ {n \atop k-1} \right]$ with the initial conditions $\left[ {0 \atop 0}\right] =1$ and $\left[ {n \atop 0}\right]= 0,\ n>0$.
