\problemdata                                                             
        {Basic Sums} % latex-friendly label for the prob.
        {sums, finding bounds} % The topic(s) of the problem content.  
        {?} % Source (if known)                                          
        {S01 Quiz 2, 5} % Usage (list ones you are aware of). 
        {Tina Wang, S02} % Last revision info (author, date).   
                                                                         
                                                                         
\begin{problem}           
\begin{problemparts}

\problempart 
What is the product of the first $n$ odd powers of two:
$\prod\limits_{k=1}^n2^{2k-1}$?   

\medskip
\solution{ 
\[
\Pi_{k=1}^n2^{2k-1} = 2^{\sum_{k=1}^n{2k-1}}
= 2^{2\sum_{k=1}^n k-\sum_{k=1}^n 1}
= 2^{n(n+1) - n}
= 2^{n^2}
\]
}
%\vspace*{2in}
\problempart What is $\sum\limits_{k=a}^b k$?   

\medskip
\solution{
\[
\sum_{k=a}^b k = 
\sum_{k=1}^b k - \sum_{k=1}^{a-1} k = 
\frac{b(b+1)}{2} - \frac{a(a-1)}{2}
\]

Also, some students computed the same answer directly
in the form
\[
\frac{(a+b)(b-a+1)}{2}
\]
by noticing that the first and last terms sum
to $(a + b)$, the second and next to last terms also
sum to $(a + b)$, etc. This means that we are adding
the term $(a + b)$ a number of times equal to
$\frac{(b - a + 1)}{2}$ sinc there are $(b -a + 1)$ terms
and we take two for each $(a + b)$.

The most common mistake on this part was to sum
up to $a$ instead of $a-1$ using the first approach.
}
%\vspace*{2in}

\problempart
Find upper and lower bounds for 
$$
\sum_{k=1}^n\frac{1}{\sqrt{k}}
$$
which differ by at most 1.    
\medskip
\solution{ 

\begin{eqnarray*}
\int_0^n \frac{1}{\sqrt{x+1}} \ dx \leq & \sum_{k=1}^n \frac{1}{\sqrt{k}} & \leq \int_0^n \frac{1}{\sqrt{x}}\ dx \\
\int_0^n \frac{1}{\sqrt{x+1}} \ dx \leq & \sum_{k=1}^n \frac{1}{\sqrt{k}} & \leq 1 + \int_1^n \frac{1}{\sqrt{x}}\ dx \\
2\sqrt{n+1} - 2 \leq & \sum_{k=1}^n \frac{1}{\sqrt{k}} & \leq 1 + 2\sqrt{n} - 2\\
2\sqrt{n+1} - 2 \leq & \sum_{k=1}^n \frac{1}{\sqrt{k}} & \leq 2\sqrt{n} - 1
\end{eqnarray*}

The most common mistake on this part of the problem was
having the wrong limits of integration (going from $1$ to $n$
instead of from $0$ to $n$). The more severe (and less common)
mistake was to switch the two functions for the upper and lower
bound.
}
\end{problemparts}

\end{problem}
                                 