For the computation of the work, we need the geometric sum formula which is \sum_{i=1}^n q^i = q*\frac{q^n-1}{q-1}.

The work of the bottom-up traversal is then computed as follows:

W_{bottom-up}(n)
= \sum_{h=1}^{\log(n)} \frac{n}{2^h}
= n * \sum_{h=1}^{\log(n)} \frac{1}{2}^h
= n * \frac{1}{2} * \frac{\frac{1}{2}^\log{n}-1}{\frac{1}{2}-1}
= n * \frac{1}{2} * \frac{\frac{1}{n}-1}{\frac{1}{2}-1}
= \frac{1}{2} * \frac{1-n}{\frac{1}{2}-1}
= \frac{1-n}{1-2}
= \frac{1-n}{-1}
= n-1

The work of the top-down traversal is computed as follows:

W_{top-down}(n)
= \sum_{h=1}^{\log(n)-1} (\frac{n}{2^{\log(n)-i}} - 1)
= (\sum_{h=1}^{\log(n)-1} \frac{n}{2^{\log(n)-i}}) - (\log(n)-1)
= (\sum_{h=1}^{\log(n)-1} \frac{n*2^i}{2^{\log(n)}}) - (\log(n)-1)
= (\sum_{h=1}^{\log(n)-1} \frac{n*2^i}{n}) - (\log(n)-1)
= (\sum_{h=1}^{\log(n)-1} 2^i) - (\log(n)-1)
= (2*\frac{2^{\log(n)-1}-1}{2-1}) - (\log(n)-1)
= (2* {2^{\log(n)-1}-1}) - (\log(n)-1)
= (2* {n/2-1}) - (\log(n)-1)
= (n - 2) - (\log(n)-1)
= n - 2 - \log(n) + 1
= n - \log(n) - 1