Expansion

Given is a product of 52 factors out of which we have to select 50 x's and 2 constants to get $x^{50}$ i.e $1(2+3..+52)+2(1+3+4..52)+..+52(1+2+..+51)$ $=\displaystyle \sum_{i=1}^n i(\small{\displaystyle \sum_{i=1}^n (i)}-i)$ $=\displaystyle \sum_{i=1}^n i\dfrac{(i(i-1))}{2}=\dfrac{1}{2}\displaystyle \sum_{i=1}^n (i^3-i^2)$ $\small( {\color{#20A900}{Using~\displaystyle \sum_{i=1}^n i=\dfrac{(n)(n+1)}{2}\\~,~\displaystyle \sum_{i=1}^n i^2=\dfrac{n(n+1)(2n+1)}{6}~and~\displaystyle \sum_{i=1}^n i^3=\dfrac{n^2(n+1)^2}{4}})}$ $=\dfrac{n(n+1)(n-1)(3n+2)}{24}$ Putting n=52, we get the desired result i.e $\Large 925327$

Using Vieta's formulas, the coefficient of $x^{50}$ is given by:

$\begin{aligned} a_{50} & = \sum_{1 \le i \le j \le 52}^{52} ij \\ & = \frac{1}{2} \left[\left(\sum_{i=1}^{52} i \right)^2 - \sum_{i=1}^{52} i^2 \right] \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Since } \left(\sum_{i=1}^{n} i \right)^2 = \sum_{i=1}^{n} i^2 + 2\sum_{1 \le i \le j \le 52}^{n} ij } \\ & = \frac{1}{2} \left[\left(\frac{(52)(53)}{2} \right)^2 - \frac{(52)(53)(105)}{6} \right] \\ & = \frac{1}{2} \left[1898884 - 48230 \right] \\ & = \boxed{925327} \end{aligned}$

Co-efficient of $x^{50} = \displaystyle \sum_{i=1}^{52}i\sum_{j≠i,j=1}^{52}j$
$\therefore Co-eff x^{50} = \dfrac{\displaystyle \sum_{i=1}^{52}i(\sum_{i=1}^{52}i-i)}{2}$
$\therefore Co-eff x^{50} = \dfrac{\displaystyle \left(\sum_{i=1}^{52}i\right)^{2} - \sum_{i=1}^{52}i^{2}}{2}$
$\therefore Co-eff x^{50} = \dfrac{ \left(\dfrac{52\cdot53}{2}\right)^{2} - \dfrac{52\cdot53\cdot105}{6}}{2} = 925327$