\problemdata      
        {?} % latex-friendly label for the prob.
        {Summations} % The topic(s) of the problem content.
        {?} % Source (if known)
        {Fall 2001, PS7; Fall 2000, Quiz 2} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).



\begin{problem}
Find a simplified closed form for each of the expressions listed below.

\bparts
\problempart $ \sum_{i=x}^{y} 2i+1 $

\solution{
There are two ways of doing this problem.  In the first, we re-form the 
sum so the lower limit is 0 instead of $x$ and apply the arithmetic sum 
formula.  

$$\sum_{i=x}^y 2i+1 = \sum_{i=0}^{y-x} 2(i+x)+1$$
$$= \sum_{i=0}^{y-x} 2i + 2x+1$$
$$= (y-x+1)(2x+1) + 2 \sum_{i=0}^{y-x} i $$
$$= (y-x+1)(2x+1) + 2 \frac{(y-x)(y-x+1)}{2} $$
$$= (y-x+1)(2x+1 + y-x) = (y-x+1)(y+x+1) = ((y+1)-x)((y+1)+x) $$
$$= (y+1)^2 - x^2.$$


In the second, we express this sum as the difference of two summations,
one involving only $y$ and one involving only $x$.  
$$
\sum_{i=x}^{y} 2i+1 = \sum_{i=0}^{y} 2i + 1 - (\sum_{i=0}^{x-1} 2i +
1).
$$
Now, since 

$$\sum_{i=0}^n 2i+1 = n+1 + 2\frac{n(n+1)}{2} = n+1 + n(n+1) = n^2 +
2n+1 = (n+1)^2,$$

we can evaluate this to

$$(y+1)^2 - x^2.$$
}

\problempart
${\displaystyle \sum_{i=0}^{\infty} \sum_{j=1}^n \left(\cfrac{j}{j+2}\right)^i}$

\solution{
First, we reorder the summations to get

$$\sum_{j=1}^n \sum_{i=0}^{\infty} \left(\frac{j}{j+2}\right)^i.$$

We can do this reordering because the value of $j$ never depends on the
value of $i$, we just use every possible pair of values for a nonnegative 
$i$ and a $j$ between 1 and $n$.

Evaluating the inner summation, which is just an infinite geometric
series, we get

$$\sum_{j=1}^n \cfrac{1}{1 - \cfrac{j}{j+2}}$$
$$= \sum_{j=1}^n \cfrac{j+2}{2}$$
$$= n + \frac{1}{2}\sum_{j=1}^n j$$
$$= n + \cfrac{n(n+1)}{4}$$

which simplifies to $\frac{1}{4}n^2 + \frac{5}{4}n$.

}

\ppart ${\displaystyle \prod_{i=1}^n 2 \cdot 4^i}$.

\solution{
We know that $\prod_{i=1}^n 2 \cdot 4^i =(2\cdot4^1) \cdot (2\cdot4^2) \cdot ... \cdot (2\cdot4^n).$
Taking $\log_2$ of both sides of the equation gives us
\begin{eqnarray*}
\log_2(\prod_{i=1}^n 2\cdot4^i)&=&\log_2(2\cdot4^1 \cdot 2\cdot4^2 \cdot ... \cdot 2\cdot4^n)\\
&=&\log_2(2\cdot4^1)+\log_2(2\cdot4^2)+...+\log_2(2\cdot4^n)\\
&=&\sum_{j=1}^n\log_2(2\cdot4^j)\\
&=&\sum_{j=1}^n(\log_2(2) + \log_2(4^j))\\
&=&\sum_{j=1}^n (1 +  2j)\\
&=&n+n(n+1)\\
&=&n^2+2n.
\end{eqnarray*}
Now, exponentiating both sides of the equation gives
$\prod_{i=1}^n 2\cdot4^i=2^{n^2+2n}$
}

\eparts
\end{problem}
