A Hundred-Row Number Pyramid

Date: 11/19/98 at 11:55:42
From: Terry Williams
Subject: Row, row, row your boat

Hello, 

I am a secondary math teacher and I am having difficulty with the 
following question.  

Given the following

        1  2
       3  4  5
      6  7  8  9
    10 11 12 13 14
   15 16 17 18 19 20

What is the last number (on the righthand side) in the one-hundredth 
row? 

Could you please send me an outline solution? I would appreciate it.

Thanks in advance,
Terry

Date: 12/01/98 at 19:54:37
From: Doctor Ameis
Subject: Re: Row, row, row your boat

Hello Terry,

Here is one approach to your problem. The approach involves setting up 
an input/output table and analyzing it to see what kind of relation 
exists between the row number (the input) and the rightmost number in a 
row (the output).

Here is the table:

   input(row number) | output(rightmost number)
   ------------------+-------------------------
        1            |       2
        2            |       5
        3            |       9
        4            |      14
        5            |      20
      100            |       ?

If you figure out the difference between consecutive output values, 
these are +3 (5 - 2), +4 (9 - 5), +5 (14 - 9), +6 (20 - 14). These 
differences increase each time by a constant amount of +1. This means 
that the relation between the input and output values is a quadratic 
function (output = a * (input)^2 + b * (input) + c  OR using more 
conventional notation: y = ax^2 + bx + c).

Pick four pairs of data (three pairs would do), construct the quadratic 
function for each pair, and solve the resulting equations for a, b, 
and c. This will give you the relation between the input and output 
that you need to determine the rightmost number in the hundredth row.

Here is one possible way to proceed:

   14 = a*(4)^2 + b*(4) + c
    2 = a*(1)^2 + b*(1) + c

Simplifying and subtracting the two equations we get:

   12 = 15a + 3b
   20 = a*(5)^2 + b*(5) + c
    9 = a*(3)^2 + b*(3) + c

Simplifying and subtracting the two equations we get:

   11 = 16a + 2b

Solving these two equations for a and b, and then using that 
information to solve for c, we get:

    a = 1/2,  b = 3/2,  c = 0

This means that one way to express the relation between input (the row 
number) and output (the rightmost number in row) is:

  output = 1/2 * (input)^2 + 3/2 * (input)

Another way to arrive at the relation is examine the input/output table 
and use "educated" trial and error to figure out the relation between 
the row number and the rightmost number. Doing it this way, you might 
see that:  

  rightmost number = 1/2 * (row number)(row number + 3).

Hope this helps.  Feel free to ask for more assistance if needed.

- Doctor Ameis, The Math Forum
  http://mathforum.org/dr.math/   

Date: 12/01/98 at 19:59:13
From: Doctor Stacey
Subject: Re: Row, row, row your boat

Hi Terry,

If you're explaining the solution to a class, I think Dr. Ameis's 
second method may be easier. It comes from realizing that the last 
number in each row (and in fact any number) is the number of numbers up 
to and including it. So you just need to count the number of numbers in 
100 rows. There are two in the first row, three in the second, ..., 101 
in the hundredth. Remember how to add 1 through 100 really quickly?  
Use the same method. First plus last, times the number of rows, divided 
by 2. 

Hope that helps!

- Doctor Stacey, The Math Forum
  http://mathforum.org/dr.math/