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\title{Assignment 1}
\author{Mike Zabrocki}
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\begin{document}
\maketitle
%\section{}
%\subsection{}

{\bf Question:}
The figure below is supposed to represent a much larger figure made of up triangles where
the base of each of the small triangles is of unit length. The figure below is 6 units on a
side, but imagine that it is one of a sequence of figures where the $n$
th figure has length $n$
units on a side.

(a) How many triangles can be found in the $r^{th}$ row from the top where $1 \leq r \leq n$?
\vskip .2in

%Well in the first row, there is one triangle,
%in the second there are three triangles
%and in the third there are 5 triangles,
%and so on....

In the $r^{th}$ row there are $2r -1$ triangles.

In the first row there is one triangle
and in every row there is the same number
of triangles as in the previous row
plus two more.  And in the beginning
(for the first row) there is 1. If the
$r^{th}$ there are $2r-1$, and the $r+1^{st}$
row has 2 more then the $r+1^{st}$ row has $2r-1+2 = 2r+1$.  Then by the principle of
mathematical induction, there are $2r-1$
triangles in the $r^{th}$ row.


\vskip .3in
(b) What is the area of the large triangle? What is the area of the large triangle divided
by the small triangle?

\vskip .2in
Heron's formula 
\begin{center}
{\tt http://www.mathwarehouse.com/geometry/triangles/area/herons-formula-triangle-area.php}
\end{center}
says that if the perimeter of the triangle is $2 S$ and the side lengths are $A,B,C$ then the area is equal to $\sqrt{S(S-A)(S-B)(S-C)}$.
In our picture the length of the side is $n$ so $A=B=C=n$ and $S = 3n/2$.  By this formula
the area of the larger triangle is equal to
$\sqrt{(3n/2)(3n/2-n)(3n/2-n)(3n/2-n)}= (\sqrt{3}/4) n^2$.  This is the area of the
larger triangle.

The area of the smaller triangle is proportional because the length of the side is $1$ instead of $n$ and so the area of each of the smaller triangles is equal to $\sqrt{3}/4$.  The area of the large triangle divided by the smaller triangle is $n^2$.

\vskip .3in
(c) Use the general figure to argue that
$$1 + 3 + 5 + \cdots + 2n-1 = n^2$$
.
\vskip .2in

The number of small triangles is equal
to $1+3+5+\cdots +(2n-1)$ and the area
occupied those triangles is going to be
$(1+3+5+\cdots +(2n-1)) \sqrt{3}/4$.
The area of the larger triangle is equal
to $(\sqrt{3}/4)n^2$.  Since the areas
of the sums of the smaller triangles
and the larger are equal, then we know
that
$$(1+3+5+\cdots +(2n-1)) \sqrt{3}/4 = (\sqrt{3}/4)n^2$$
and so
$$(1+3+5+\cdots +(2n-1)) = n^2~.$$


\vskip .5in
(1) look for simarlities and patterns

(2) guess and check and alter (if it doesn't work)

(2.5) generate data, do examples

(3) check the differences and the second differences

For example: our sequence increases like $1,4,9,16,25$ and the first differences are $1, 3,5,7,9$ the fact that those increase linearly means that the formula for the sequence itself will have a square.

(4) look for patterns, generalize

(5) rewrite your question in different language

(6) look it up (roughly, this means ask Google)

(7) ask someone who knows the answer

(8) learn more math


\vskip .3in
{\bf Problem}:
Show that $n^2+n+41$ is prime for $n \geq 1$


\begin{center}
    \begin{tabular}{c|c|c}
    $n=$&$n^2+n$&$n^2+n+41$\cr
    \hline
    1&2&43\cr
    2&6&47\cr
    3&12&53\cr
    4&20&61\cr
    5&30&71\cr
    6&42&83\cr
    7&56&97\cr
    8&72&113\cr
    $\vdots$&&\cr
    40&$40\cdot41$&$40\cdot41+41= 41^2$
    \end{tabular}
\end{center}


\end{document}