Which one of the following are correct? 
Let, $A=   \left[ {\begin{array}{cc}    -1 & 2 \\    0 & -1 \\  
 \end{array} } \right]$ , and $B = A + A^2 + A^3 +···+ A^{50}$. Then
$(A) B^2 = I $
$(B) B^2 = 0$ 
$(C) B^2 = A$ 
$(D) B^2 = B$

Eigenvalues of $A$ are $-1,-1$. So, Eigenvalues of $B$ are $0,0$. So, $\det(B)=0$. So, options (A) and (C) can be eliminated. How do I eliminate further?. By Cayley-Hamilton Theorem $A^2+2A+I=0$. I found $A^3,A^4,...$ using Cayley-Hamilton Theorem. It was time-consuming. Given matrix is not diagonalizable. Can you please help me?

Since $A-I$ is triangular with nonzero diagonal entries, it follows that $A-I$ has nonzero determinant, so is invertible, hence $I-A$ is also invertible.

Note also that any two poynomials in $A$ commute.

\begin{align*}
\text{Then}\;\;(I-A)B&=(I-A)(A+A^2+A^3+\cdots+A^{50})\\[4pt]
&=(A + A^2 + A^3 +···+ A^{50})-(A^2 + A^2 + A^3 +···+ A^{51})\\[4pt]
&=A-A^{51}\\[4pt]
&=(I-A^{50})A\\[4pt]
&=(I+A^{25})(I-A^{25})A\\[4pt]
&=(I+A)(1-A+A^2-\cdots+A^{24})(I-A^{25})A\\[4pt]
&=(I+A)C,\;\text{where}\;C=(1-A+A^2-\cdots+A^{24})(I-A^{25})A\\[4pt]
\implies\;\bigl((A-I)B\bigr)^2&=\bigl((I+A)C\bigr)^2\\[4pt]
\implies\;(A-I)^2B^2&=(I+A)^2C^2\qquad\text{[since $B,C$ commute with all polynomials in $A$]}\\[4pt]
\implies\;(A-I)^2B^2&=0\qquad\text{[since $(I+A)^2=A^2+2A+I=0$]}\\[4pt]
\implies\;B^2&=0\qquad\text{[since $(I-A)$ is invertible]}\\[4pt]
\end{align*}
