$A=\begin{pmatrix} -1 &2 \\ 0& -1 \end{pmatrix}$
$A^{2}=\begin{pmatrix} -1 &2 \\ 0& -1 \end{pmatrix}\begin{pmatrix} -1 &2 \\ 0& -1 \end{pmatrix}=\begin{pmatrix} 1 &-4 \\ 0& 1 \end{pmatrix}$
$A^{3}=\begin{pmatrix} -1 &6 \\ 0& -1 \end{pmatrix}$
$A^{4}=\begin{pmatrix} 1 &-8 \\ 0& 1 \end{pmatrix}$
Given,$B=A+A^{2}+A^{3}+......+A^{50}$
$B=\begin{pmatrix} -1+1-1+1....+1 \ \ \ \ \ \ \ &2-4+6-8+10+.......-100 \\ 0& -1+1-1+1....+1 \end{pmatrix}$
$B=\begin{pmatrix} 0 \ \ &2-4+6-8+10+.......-100 \\ 0& 0 \end{pmatrix}$
Let $X=2-4+6-8+10+.......-100$
We can divide it into 2 series
$2+6+10+...98$ and $-4-8-12.......-94-100$
Let's take $2+6+10+...98$
This sequence is an Arithmetic Progression.
Sum of Arithmetic Progression=$\frac{n}{2}[2a+(n-1)d]$
Here a=2 , d=4 , n=25
$\frac{25}{2}[4+24 \times 4]$
$=1250$
Now take $-4-8-12.......-94-100$
$-4-8-12.......-94-100$ = $-4(1+2+3+....+25)$
=$-4 \times\frac{25 \times 26}{2}$
=$-1300$
$X=2-4+6-8+10+.......-100$
$X=2+6+10+...+98-4-8-12...-100 \\ X=1250-1300 \\ X=-50$
Put the value of X into the matrix
$B=\begin{pmatrix} 0 &-50 \\ 0 & 0 \end{pmatrix}$
$B^{2}=\begin{pmatrix} 0 &-50 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 &-50 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 &0 \\ 0 & 0 \end{pmatrix}$
$B^{2}=0$
Hence,Option$(B)$$B^{2}=0$ is the correct choice.