Dat Do edited beginproblem_a_You_a.tex  about 10 years ago

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\smallskip  \noindent  \\  (b) Since we are still receiving one extra coupon for each consecutive day, $\sum\limits_{i=1}^n i$ can still represent the number of coupons received. However, the value of the coupons is: $1 + 2 + 4 + \ldots+ (n-1) + 2(n-1)$ which can be represented as $\sum\limits_{i=1}^n 2^{n-i}$.   \smallskip  We must multiply the quantity and the value together during each sum to get the total value resulting in: $W(n) = \sum\limits_{i=1}^n i*2^{n-i}$  \smallskip  \noindent  \\  (c) For $n = 1$ through $n = 6$,   \begin{itemize}  \item $W(1) = 1$