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Dat Do edited beginproblem_a_You_a.tex
about 10 years ago
Commit id: 762b909dd1ef033245db54b976e9389fcff9e5b0
deletions | additions
diff --git a/beginproblem_a_You_a.tex b/beginproblem_a_You_a.tex
index 166c24b..e2ddf33 100644
--- a/beginproblem_a_You_a.tex
+++ b/beginproblem_a_You_a.tex
...
\item $W(4) = 26$
\item $W(5) = 57$
\end{itemize}
We know that the closed form expression should involve a value slightly larger than $2^n$, so we
try try: $2^{n+1}$ and get \begin{itemize}
\item $n = 1$ as $4$
\item $n = 2$ as $8$
\item $n = 3$ as $16$
...
\item $n = 5$ as $64$
\end{itemize}
The
summation can difference between $2^{n+1}$ and $W(n)$ appears to be
represented in -n-2 for each term, so the closed form
as so: $W(n) = \sum\limits_{i=1}^n expression for $W(n)$ must be $\sum\limits_{i=1}^n i*2^{n-i} = 2^{n+1}-n-2$
\begin{problem}