MAT 118 Spring 2013

- Row $i$ has length $i+1$ and starts and ends with a $1$.
- Each interior element is the sum of the two above it.

The standard notation for the number in row $n$ and column $k$ is $\left ( \begin{array}{c} n\\k \end{array} \right )$. So $\left ( \begin{array}{c} n\\0 \end{array} \right )$ is always $1$, and $\left ( \begin{array}{c} n\\n \end{array} \right )$ is always $1$, whereas $\left ( \begin{array}{c} 5\\3 \end{array} \right ) = 10$.

The entries in Pascal's triangle appear in the *Binomial Theorem*:
$$(a+b)^n = \left ( \begin{array}{c} n\\0 \end{array} \right ) a^n +
\left ( \begin{array}{c} n\\1 \end{array} \right ) a^{n-1}b +
\left ( \begin{array}{c} n\\2 \end{array} \right ) a^{n-2}b^2 + \dots +
\left ( \begin{array}{c} n\\n-2 \end{array} \right ) a^2b^{n-2}+
\left ( \begin{array}{c} n\\n-1 \end{array} \right ) ab^{n-1} +
\left ( \begin{array}{c} n\\n \end{array} \right ) b^n.$$

So for example
$$(a+b)^0 = 1$$
$$(a+b)^1 = a+b$$
$$(a+b)^2 = a^2 + 2ab + b^2$$
$$\dots$$
$$(a+b)^5 = a^2 + 5ab^4 + 10a^2b^3 + 10 a^3b^2 + 5a^4b + b^5.$$

You can check that
$$(1+1)^5 = 1 + 5 + 10 + 10 + 5 + 1$$
or $2^5 = 32$. More generally, a similar calculation shows that *the
sum of the numbers in row $n$ is $2^n$.* Make sure you understand why
this follows from the Binomial Theorem.

$1$

$1$$1$

$1$$0$$1$

$1$$1$$1$$1$

$1$$0$$0$$0$$1$

$1$$1$$0$$0$$1$$1$

$1$$0$$1$$0$$1$$0$$1$

$1$$1$$1$$1$$1$$1$$1$$1$

$1$$0$$0$$0$$0$$0$$0$$0$$1$

$1$$1$$0$$0$$0$$0$$0$$0$$1$$1$

$1$$0$$1$$0$$0$$0$$0$$0$$1$$0$$1$

$1$$1$$1$$1$$0$$0$$0$$0$$1$$1$$1$$1$

$1$$0$$0$$0$$1$$0$$0$$0$$1$$0$$0$$0$$1$

$1$$1$$0$$0$$1$$1$$0$$0$$1$$1$$0$$0$$1$$1$

$1$$0$$1$$0$$1$$0$$1$$0$$1$$0$$1$$0$$1$$0$$1$

$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$$1$

etc.

Notice that the triangle formed by the first 16 rows (number 0 to 15) contains three smaller triangles of height 8 and of the form:

$1$

$1$$1$

$1$$0$$1$

$1$$1$$1$$1$

$1$$0$$0$$0$$1$

$1$$1$$0$$0$$1$$1$

$1$$0$$1$$0$$1$$0$$1$

$1$$1$$1$$1$$1$$1$$1$$1$

and that each of them contains three triangles of height 4 and of the form:

$1$

$1$$1$

$1$$0$$1$

$1$$1$$1$$1$

and that each of those contains three triangles of height 2 and of the form:

$1$

$1$$1$

and that each of them contains three "triangles" of height 1!

1

11

121

1001

11011

121121

1002001

11022011

121212121

1000000001

etc.

Note that the entries in this table can be calculated directly
using addition *modulo* 3:
\begin{array}{c|ccc}
+&0&1&2\\\hline
0&0&1&2\\
1 &1&2&0\\
2&2&0&1
\end{array}
Can you see a Sierpinski-like process here? Explain.
Tell as much about it as you can. Feel free to calculate additional rows!

Check that this definition also gives $\left ( \begin{array}{c} 5\\3 \end{array} \right ) = 10$.

Show that the coefficients defined using this definition satisfy: $$\left ( \begin{array}{c} n+1\\k+1 \end{array} \right ) = \left ( \begin{array}{c} n\\k \end{array} \right ) + \left ( \begin{array}{c} n\\k+1 \end{array} \right ).$$ Hint: put the terms on a common denominator.

Remember: Collaboration is fine, but
what you hand in *must be in your own words.* Handing in
something you copied is plagiarism and will cost you if it is
detected. Write down what you tried and how it worked.

Hand in your work at the recitation meeting (Mon, Wed or Thur)
next week.