Time for MATH MINUTE! (provide your favorite theme
music here).
Get
out your paper & pencil because I have a new puzzle for you!!
Puzzle
#47: PASCAL TRIANGLE
The Pascal
triangle is formed by starting with the number 1 at the top, followed with a
second row of two 1's. To get the third row start with a 1, then add the two
numbers 1+1 from above in the second row to get 2, and end with another 1. The
3's in row four come from 1+2 and 2+1 in row three above. Repeat this formula
to get the next row, and so on. Each row always starts with 1 and ends with 1.
This is
called the Pascal triangle, even though it has no bottom. It has many
interesting properties.
A. Find the
next two rows of this triangle.
Notice the tree-like
form, except there is no bottom?
If you drew
a line down the center from 1 to 2 to 6 to 20 and so on you can see that the
left side looks like a mirror image of the right side of this tree/triangle.
This is called symmetry about the center line.
B. Except
for the first number in the center line (1) the other numbers all appear to be
even numbers (divisible by 2). Is this always true? Explain why??
Notice that
the two triangles in the second Pascal triangle below which form a 6-point
star* around the number 4 in row 5 satisfy the property: if you multiply the 3
numbers of one triangle and the 3 numbers of the other triangle you get the
same result, 1x6x5 = 30 = 3x1x10?
C. Does
this property hold for all of the other such stars (notice we can not form a
star around a 1)? Check it out on some other numbers.
D. I claim
that this property is true for all such stars about a number other than 1 in
Pascal's triangle. Can you explain why?
Hint: The
kth number in the nth row of Pascal's Triangle is the combination of n things
taken (k-1) at a time. The combination of n things taken k at a time is defined
by C(n,k) = n!/[(k)!(n-k)!], for k=0, . . . , n, where n! = 1x2x. . . x(n-1)xn,
and 0! = 1 by definition.
Then the
Pascal triangle satisfies the following property of combinations by definition:
C(n,k) = C(n-1,k-1) + C(n-1,k), for k>0.
*This
is puzzle PoW1088 from the Macalester College POW.
Send
your comments, ideas and solutions before Monday to this email address, and be sure to put
MM in the subject line: parksjm@potsdam.edu
Visit
us here online to see the results every Monday.
See
you next time on MATH MINUTE! (theme music fades out here).