this is for holding javascript data
Dat Do edited beginproblem_a_You_a.tex
about 10 years ago
Commit id: 6080a5b5d191b7dbaab95456400a5e63e0980a6b
deletions | additions
diff --git a/beginproblem_a_You_a.tex b/beginproblem_a_You_a.tex
index 8d00543..258dbb1 100644
--- a/beginproblem_a_You_a.tex
+++ b/beginproblem_a_You_a.tex
...
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 $W(1) = 1$.
...
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$
\blacksquare \end{proof}