But where's the remaining information?

A quadratic( $ax^2+bx+c=0$ ) has only one root when its discriminant is zero( i.e $b^2=4ac$ ). Thus:-

$(\not2(-1)^{i+1}x_{i+1})^2=\not4 x_ix_{i+2}$ $\implies x_{i+1}^2=x_ix_{i+2}$

$\implies x_{i},x_{i+1},x_{i+2}$ are in GP $\forall i\in[1,n-2]$ . Hence $x_1,x_2,x_3,\cdots , x_n$ form a GP such that $\dfrac{x_{i+1}}{x_i}=\text{Constant}=\dfrac{x_2}{x_1}\implies\dfrac{x_1x_{k+1}}{x_kx_2}=1$

Hence summation is simply:-

$\Large\displaystyle\sum_{n=2}^{51}\sum_{k=1}^{n-1}(1)$

$=\Large\displaystyle\sum_{n=2}^{51}(n-1)$ $\large =\dfrac{50(51)}{2}=\boxed{1275}$

For the quadratic equation in $y$ to have only one root, the discriminant ( $b^2-4ac$ ) is equal to zero, that is:

$\begin{aligned} \left(2(-1)^{i+1}x_{i+1}\right)^2 - 4x_i x_{i+2} & = 0 \\ \implies 4x_{i+1}^2 & = 4x_i x_{i+2} \\ x_{i+1}^2 & = x_i x_{i+2} \\ \frac {x_{i+1}}{x_i} & = \frac {x_{i+2}}{x_{i+1}} \\ \implies \frac {x_{i+1}}{x_i} & = \text{constant} = \frac {x_2}{x_1} \end{aligned}$

Now, we have:

$\begin{aligned} \sum_{n=2}^{51} \sum_{k=1}^{n-1} \frac{x_1\color{#3D99F6}{x_{k+1}}}{\color{#3D99F6}{x_{k}}x_2} & = \sum_{n=2}^{51} \sum_{k=1}^{n-1} \frac{x_1\color{#3D99F6}{x_{2}}}{\color{#3D99F6}{x_{1}}x_2} \\ & = \sum_{n=2}^{51} \sum_{k=1}^{n-1} 1 \\ & = \sum_{n=2}^{51} (n-1) \\ & = \sum_{n=1}^{50} n \\ & = \frac {50(51)}2 = \boxed{1275} \end{aligned}$