Extension for Perfect Fit
If we
extend the Perfect Fit problem to look at the surface area of Cuisenaire rods
placed so as to create an “L” we can see that a general formulas for the
surface area is
,
where n represents the number of rods being used and h represents
the height of the rods.

Looking at
the pattern being develop one can see that the surface area of the top and
bottom will be 2 times the number of rods being used. (This is true since each of the rods have a
length and width value of one.) The
lateral surface area becomes the number of sides plus 1 times 2. If the height of the rods change then the
lateral surface area will change by a factor of the height. Thus if the “L” has a height of “h” units, the total surface will become
. The lateral surface area can be found the
same way as if we were using a
rod,
but we would need to multiply that area by the number of
cubes
that would make up the height.
The picture at left represents the side view of an “L”
created by the Cuisenaire rods.
When using the
rods,
the lateral surface would be represented by the length times the width by
the number of rods making up the length of the “L”. As the height of the “L” increases, the
lateral surface will increase by the number of
cubes
which would be needed to create the
rod.
If we take the Cuisenaire rods and form an “X”, a general equation for finding
the total surface area of the figure formed would be
. In order to see this
examine the figures shown below.

Examining the cases above and others, the total surface area of the “X’s” , becomes
5 rods ![]()
9 rods ![]()
13
rods ![]()
therefore if it took q rods
to create the pattern the generalized formula becomes
. Since it takes at
least 5 rods to create the first pattern and then 4 rods gets added each time
we extend the pattern after that, the q becomes
,
where n $1. If we substitute
for q. The generalized formula
becomes
. When simplified the formula becomes
When we use a different length rod from the
,
the generalized formula would become
. Where h represents the height of the
rod. Thus, the general form for finding
the surface area of Cuisenaire rods
placed to form an “X” becomes
.
If
we arrange the rods in the form of a “T”.
The generalized equation of the surface area becomes
,
where n is the number of rods needed to create the “T” and h represents the
height of the rod.
If
we examine the pattern in the total surface areas of the figures above we
arrive at the formula
as long as the rods
are
. If we change the
height of the rods then the formula becomes
which simplifies to
.
Finally,
if we arrange the rods to become rectangular solids with a square base, the
generalized formula for the total surface area becomes
,
where n represents the number of
rods on a side and h represents the height of the rods.
In
examining the pattern shown in the box below, we see that if the rods used are
the total surface area becomes
1 rod used 2(1) + 4(1)
4 rods used 2(4) + 4(2)
9 rods used 2(9) + 4(3)

Thus
the general equation for rectangular solids with square bases composed of
rods becomes
. If we increase the height of the rods. Then
the general formula becomes
which simplifies to ![]()
Extension of the Train Problem
I’m
not sure but I think we may have done the extension in class. In class we looked at the number of ways of
choosing sub-collections of k objects from an initial set of n
objects. To me this sounds like the
extension of the Train Problem, “How
many trains of length n can have k pieces?” If this is the same problem then the solution
becomes
.

Mathematical Induction
It has been several years since I have done any proofs using mathematical induction. I remember that the proofs start with let n = 1 and works its way down to letting n = k. I also remember that they were also some of the hardest proofs to do and they really did not make a great deal of sense to me.
Example:
Prove
that for all natural numbers n,
.
Let n =1
Then 1 + 2 + 3 + 4 + . . . + n is actually just 1
becomes
when
you replace n with 1.
![]()
Thus the
original statement
is true when n =1.
Let n = k
Assume that
is true.
Let n = k + 1
Then 1 + 2 + 3 + 4 + . . . + k + k + 1 = [1 + 2 + 3 + 4 + . . . + k] + k + 1
![]()
![]()
![]()
![]()
therefore if n = k, ![]()