Let A = [ cases 1 & (1)/(51) 0 & 1 cases ] . If B = [ cases 1 & 2 -...

Given,

A = \left[\begin{cases}1 & \dfrac{1}{51} \\ 0 & 1\end{cases}\right],

B = \left[\begin{cases}1 & 2 \\ - 1 & - 1\end{cases}\right]A\left[\begin{cases}- 1 & - 2 \\ 1 & 1\end{cases}\right]

So, $B = MAN$

Now let M = \left[\begin{cases}1 & 2 \\ - 1 & - 1\end{cases}\right] \text{ and } N = \left[\begin{cases}- 1 & - 2 \\ 1 & 1\end{cases}\right]

Now solving,

\overset{\left(M\right)}{\left[\begin{cases}1 & 2 \\ - 1 & - 1\end{cases}\right]}\overset{\left(N\right)}{\left[\begin{cases}- 1 & - 2 \\ 1 & 1\end{cases}\right]} = \left[\begin{cases}1 & 0 \\ 0 & 1\end{cases}\right]

$∴ MN = I = NM$

Now using, $B = MAN$ we get

${B}^{n} = {\left(MAN\right)}^{n} = \left(MAN\right)\left(MAN\right) . . . . . . . \left(MAN\right)$

$⇒ {B}^{n} = MA\mathop{\mathop{NM}_{⏟}}_{I}A\mathop{\mathop{NM}_{⏟}}_{I} . . . . . . . \mathop{\mathop{NM}_{⏟}}_{I}AN$

$⇒ {B}^{n} = M\mathop{\mathop{AAAAA . . . . . A}_{⏟}}_{n\text{times}}N = M{A}^{n}N$

Now, using A = \left[\begin{cases}1 & \dfrac{1}{51} \\ 0 & 1\end{cases}\right]

⇒ A = \left[\begin{cases}1 & 0 \\ 0 & 1\end{cases}\right] + \left[\begin{cases}0 & \dfrac{1}{51} \\ 0 & 0\end{cases}\right] = I + E

Now finding, {E}^{2} = \left[\begin{cases}0 & \dfrac{1}{51} \\ 0 & 0\end{cases}\right]\left[\begin{cases}0 & \dfrac{1}{51} \\ 0 & 0\end{cases}\right] = \left[\begin{cases}0 & 0 \\ 0 & 0\end{cases}\right]

$⇒ {E}^{2} = 0$, so all higher power will also be zero,

$∴ {A}^{n} = {\left(I + E\right)}^{n} = I + nE + \mathop{\mathop{{}^{n}{C}_{2}{E}^{2}}_{⏝}}_{0} + \mathop{\mathop{{}^{n}{C}_{3}{E}^{3}}_{⏟}}_{0} + \mathop{\mathop{{}^{n}{C}_{4}{E}^{4}}_{⏟}}_{0} . . . . . . . . . . .$

$⇒ {A}^{n} = I + nE$

⇒ {A}^{n} = \left[\begin{cases}1 & \dfrac{n}{51} \\ 0 & 1\end{cases}\right]

So, {B}^{n} = M{A}^{n}N = \left[\begin{cases}1 & 2 \\ - 1 & - 1\end{cases}\right]\left[\begin{cases}1 & \dfrac{n}{51} \\ 0 & 1\end{cases}\right]\left[$\begin{cases}- 1 & - 2 \\ 1 & 1\end{cases}$\right]$$⇒ {B}^{n} =$\left[$\begin{cases}1 & \dfrac{n}{51} + 2 \\ - 1 & \dfrac{- n}{51} - 1\end{cases}$\right]\left[$\begin{cases}- 1 & - 2 \\ 1 & 1\end{cases}$\right]$</p><p>$⇒ {B}^{n} = \left[$\begin{cases}1 + \dfrac{n}{51} & \dfrac{n}{51} \\ \dfrac{- n}{51} & 1 - \dfrac{n}{51}\end{cases}$\right]$</p><p>$\mathop{∑}_{n = 1}^{50}{B}^{n} = \left[$\begin{cases}50 + \dfrac{50 \cdot 51}{2 \cdot 51} & \dfrac{50 \cdot 51}{2 \cdot 51} \\ \dfrac{- 50 \cdot 51}{2 \cdot 51} & 50 - \dfrac{50 \cdot 51}{2 \cdot 51}\end{cases}$\right] = \left[$\begin{cases}75 & 25 \\ - 25 & 25\end{cases}$\right]$</p><p>$∴$Sum$= 100$