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,