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\headers{Math378}{Quiz 1}{Fri. Jan 23, 2026}
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\hfill Name: \underline{\hspace{3.5in}} 

\noindent
\textbf{Directions:} Show all work.  

\begin{enumerate}

\item\points{6} For $n\ge 0$, let 
\[S_n = \sum_{k=1}^{2n} (-1)^k k^2 = -1^2 + 2^2 - 3^2 + 4^2 - \cdots - (2n-1)^2 + (2n)^2.\]  
Prove that for each non-negative integer $n$, we have that $S_n = (2n+1)n$.
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\item\ppoints{2}{2} Recall that the adjusted Fibonacci sequence is defined by $\hat F_0 = \hat F_1 = 1$ and $\hat F_n = \hat F_{n-1} + \hat F_{n-2}$ for $n\ge 2$.
\begin{enumerate}
	\item Let $n$ be a positive integer and let $t$ be the maximum integer such that $\hat F_t \le n$.  Prove that $n-\hat F_t < \hat F_{t-1}$.  (Hint: there is a short proof, without induction/no minimum counter-example.  If stuck, then try a proof by contradiction.)
	\vspace{7cm}
	
	\item Prove that each positive integer $n$ is a sum of distinct, non-consecutive adjusted Fibonacci numbers.  For example, if $n=20$, then we have $20 = 13 + 5 + 2 = \hat F_6 + \hat F_4 + \hat F_2$.
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