deletions | additions       diff --git a/beginproblem_a_You_a.tex b/beginproblem_a_You_a.tex  index fed93cb..8d00543 100644  --- a/beginproblem_a_You_a.tex  +++ b/beginproblem_a_You_a.tex 
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The difference between $2^{n+1}$ and $W(n)$ appears to be $-n-2$ for each term, so the closed form expression for $W(n)$ must be $\sum\limits_{i=1}^n i*2^{n-i} = 2^{n+1}-n-2$. We can prove that this expression is correct using induction.  \\   \begin{proof}  \textbf{Proof:}  \\*  Base Case: For $W(1)$, $2^{1+1}-1-2 = 1$ which is correct since $W(1) = 1$. 
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We see that each term $W(k+1) = 2W(k) + (k+1)$. so,  \\$W(k+1) = 2(2^{k+1}-k-2) + (k+1)$  \\$W(k+1) = 2^{k+2}-2k-4+k+1$  \\$W(k+1) = 2^{k+2}-k-3$ \qed \blacksquare