Given $A=\begin{bmatrix} -1 & 2\\ 0 & -1 \end{bmatrix}$
$A^2=\begin{bmatrix} 1 & -4\\ 0 & 1 \end{bmatrix}$
$A^3=\begin{bmatrix} -1 & 6\\ 0 & -1 \end{bmatrix}$
$A^4=\begin{bmatrix} 1 & -8\\ 0 & 1 \end{bmatrix}$
$A^5=\begin{bmatrix} -1 & 10\\ 0 & -1 \end{bmatrix}$
so we can generalize $A^k=$$\begin{bmatrix} (-1)^k & 2k.(-1)^{k+1}\\ 0 & (-1)^k \end{bmatrix}$
Given matrix A=$\begin{bmatrix} a_{11} &a_{12} \\ a_{21} & a_{22} \end{bmatrix}$
When we sum up $A+A^2+A^3+....A^{50}=B$ it is very clear that $b_{21}=0$ and since half of the matrix powers are odd and half are even, so when we add all of them together then $b_{11}=b_{22}=0$
Now, we need to compute the entry $b_{12}$
To compute it I take sum of all entries of $a_{12}$ in $A+A^3+A^5+...A^{49}$
So this will be like $2+6+10+....+98 = \frac{98+2}{2} *25 =1250$ and this will be positive in sign.
Now, I take the sum of all entries of $a_{12}$ in $A^2+A^4+....A^{50}$
and this is like $-4-8.....-100 = \frac{-104}{2} * 25=-1300$
So, $b_{12}$ entry would be $1250-1300=-50$
Hence our matrix B would be $\begin{bmatrix} 0 &-50\\ 0 & 0 \end{bmatrix}$
$B^2=0.$ option(B)
Note : To take sum of AP $a_1+a_2+a_3+...a_n= \frac{a_n(last\, term)+a_1 (first \, term)}{2}* \text{number of terms}$