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\begin{document}

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  \firstpageheader{ Math 15A }{ Homework 3 }{ February 1, 2019 }
  \runningheader{ Math 15A }{ Homework 3, Page \thepage\ of \numpages}{ February 1, 2019 }

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    \makebox[0.4\textwidth]{Name: \enspace\hrulefill}

    \vspace{0.2in}
    \makebox[0.4\textwidth]{Pid:\enspace\hrulefill}
  \end{flushright}

  \begin{questions}
    \question[10]
      Let $a_n$ be a sequence such that $a_1 = 9$, $a_2 = 41$, and
			$a_{n + 2} = 9a_{n + 1} - 20a_n$. Show that $a_n = 4^n + 5^n$.

      \begin{solution}[\stretch{1}]
      \end{solution}
      \newpage
    \question
      We say that $L$ is a list of powers of $x$ iff
			\begin{itemize}
			  \item either $L = x^k$ for some positive integer $k$ or
			  \item $L = (x^k, L')$ where $L'$ is a list of powers of $x$ and
			    $k$ is a positive integer.
			\end{itemize}
			
			Let $L$ be a list of powers of $x$. We say that the sum of $L$
			with $x = v$ denoted by $\sum L\big\rvert_{x = v}$
			\begin{itemize}
			  \item is equal to $v^k$ whether $L = x^k$ and
			  \item is equal to $v^k + \sum L'\big\rvert_{x = v}$ whether
			    $L = (x^k, L')$.
			\end{itemize}
			
			Prove that for any list $L$ of powers of $x$ there is a polynomial such that
			$\sum L\big\rvert_{x = v} = p(v)$ for all real numbers $v$.

      \begin{solution}[\stretch{1}]
      \end{solution}
      \newpage
    \question[10]
      Prove that $\sum\limits_{i = 1}^n (i + 1) 2^i = n 2^{n + 1}$ for all integers
			$n \ge 1$.

      \begin{solution}[\stretch{1}]
      \end{solution}
      \newpage
  \end{questions}
\end{document}