$DC =\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]= I$
$\begin{aligned} & B = CAD \\ & B ^{ n }=\underbrace{( CAD )( CAD )( CAD ) \ldots( CAD )}_{ n -\text { times }}\end{aligned}$
$\Rightarrow B ^{ n }= CA ^{ n } D$
$A ^2=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & \frac{2}{51} \\ 0 & 1\end{array}\right]$
$A^3=\left[\begin{array}{cc}1 & \frac{3}{51} \\ 0 & 1\end{array}\right]$
similarly $A^{ n }=\left[\begin{array}{cc}1 & \frac{ n }{51} \\ 0 & 1\end{array}\right]$
$B ^{ n }=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]\left[\begin{array}{cc}1 & \frac{ n }{51} \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$
$=\left[\begin{array}{cc}1 & \frac{ n }{51}+2 \\ -1 & -\frac{ n }{51}-1\end{array}\right]\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right]$
$=\left[\begin{array}{cc}\frac{ n }{51}+1 & \frac{ n }{51} \\ -\frac{ n }{51} & 1-\frac{ n }{51}\end{array}\right]$
$\sum \limits_{ n =1}^{50} B ^{ n }=\left[\begin{array}{cc}25+50 & 25 \\ -25 & -25+50\end{array}\right]=\left[\begin{array}{cc}75 & 25 \\ -25 & 25\end{array}\right]$
Sum of the elements $=100$
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$1 + 6 + \frac{{9({1^2} + {2^2} + {3^2})}}{7} + \frac{{12({1^2} + {2^2} + {3^2} + {4^2})}}{9} + \frac{{15({1^2} + {2^2} + .... + {5^2})}}{{11}} + ...$ up to $15$ terms, is
If $\sum\limits_{n=1}^{100} a_{2 n+1}=200$ and $\sum\limits_{n=1}^{100} a_{2 n}=100,$ then $\sum\limits_{n=1}^{200} a_{n}$ is equal to