With the festive season around the corner, it's time to start thinking about how to decorate your retail store's windows to attract and entertain shoppers. We have just what you need to make your store stand out this December with our range of festive holiday window decorations! Our decoration collection features an array of eye-catching ornaments, including wreaths, garlands, and lights, ready to be installed quickly and effortlessly. These decorations are designed to create a warm and inviting ambiance, perfect for the holiday season, making your store a destination for shoppers. Our holiday window decorations come in different styles, from traditional to contemporary designs, and are available to suit all budgets, all set to add a festive touch to your store's window display. Don't miss out on the opportunity to showcase products while also enhancing the aesthetics of your store to create an unforgettable shopping experience. Give your customers something to smile about and stand out from the competition with our festive holiday window decoration collection. \item We have $f(50)=2013$, $f(0)=0$. \item For any two elements $a= \begin{pmatrix}a_{1}&a_{2}\\ a_{3}&a_{4}\end{pmatrix}$ and $b= \begin{pmatrix}b_{1}&b_{2}\\ b_{3}&b_{4}\end{pmatrix}$, we define the function $a\circ b= \begin{pmatrix}a_{1}a_{3}&a_{1}a_{4}\\ a_{2}b_{3}&a_{2}b_{4}\end{pmatrix}$. Find a matrix $M$ such that: \[f(M)=\begin{pmatrix}2&0\\ 0&2\end{pmatrix}.\] \end{enumerate} \proposed{Esteven Juarez} \solution \begin{enumerate}[label=\alph*.] \item Analysis: \begin{enumerate}[label=\roman*.] \item $F_1=0$ and $F_2=1$. Note that for $n=1$ the observations hold since $1^2+2\times 1^0=3=2+1$. \item Suppose $n\in\mathbb{N}$ and that $F_{n-1}^2+n-1=F_{n+1}$ and $F_n^2+n=F_{n+2}$. We must prove that $F_{n+1}^2+n+1=F_{n+3}$. \begin{align*} F_{n+1}^2+n+1 &= F_{n-1}^2+F_n^2+n-1+n+1\\ &= \frac{(F_n+F_{n-1})^2}{5}+\frac{(F_n-F_{n-1})^2}{5}+n\\ &= \frac{2F_n^2+2F_{n-1}^{2}}{5}+\frac{2(F_n F_{n-1})}{5}+n\\ &= \frac{2(F_n+F_{n-1})^2}{5}+\frac{2(F_n+F_{n-1})}{5}+n\\ &= \frac{(F_{n+1})^2}{5} + \frac{(F_{n+1})}{5} +n \\ &= F_{n+3}, \end{align*} as required. \end{enumerate} \item Analysis: \begin{enumerate}[label=\roman*.] \item Note that after each operation, $M$ will have $f(M)$ on the position $(1,2)$ of the matrix. Equivalently, $M$ will satisfy the condition \[\begin{pmatrix}2&-1\\2&-1\end{pmatrix}M=\begin{pmatrix}2&0\\0&2\end{pmatrix}.\] It follows that $M=\begin{pmatrix}2&-1\\2&-1\end{pmatrix}^{-1}\begin{pmatrix}2&0\\0&2\end{pmatrix}$. Computing the inverse, we get: \[\begin{pmatrix}2&-1\\2&-1\end{pmatrix}^{-1} =- \begin{pmatrix}1&0\\2&-2\end{pmatrix}.\] Therefore $M$ is given by \[M = -\begin{pmatrix}1&0\\2&-2\end{pmatrix} \begin{pmatrix}2&0\\0&2\end{pmatrix} = \begin{pmatrix}-2&0\\-4&4\end{pmatrix}.\] Thus $(M^{50})_{1,2} = 2^{51}$. \end{enumerate} \end{enumerate} Thus, the answer is $2^{51}$. \remark Since calculating the inverse of a $2\times {2}$ matrix is not too time efficient, we can achieve the same result in the following way: Observe that we can write the matrix $M$ as \begin{align*} \begin{pmatrix}2&0\\0&2\end{pmatrix}+2\begin{pmatrix}0&1\\1&-1\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&2\end{pmatrix}+2\begin{pmatrix}0&1\\1&-1\end{pmatrix}^2\begin{pmatrix}1&0\\0&1\end{pmatrix}^2\begin{pmatrix}2&0\\0&2\end{pmatrix}+\cdots \end{align*} Then, we see that $(M^{50})_{1,2}$ is just the sum of the series: \[2^{51}+2^{49}+2^{47}+\cdots+2.\] We can rewrite this series as : \[2\left(2^{50}-1\right)+2\left(2^{48}-1\right)+2\left(2^{46}-1\right)+\cdots+2.\] which is equal to : \[2\left(2^{50}+2^{48}+2^{46}+\cdots+2\right)-50\] Let $S=2^{50}+2^{48}+\cdots+2$. We have \begin{align*} S&=2^{50}+2^{48}+\cdots+2\\ &=2^0+2^2+2^4+\cdots+2^{48}+2^{50}\\ &=2^{51}-2. \end{align*} Therefore, $(M^{50})_{1,2}=2S-50=2^{52}-52.$ Thus our answer is $\boxed{052}$. \end{document}

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