posted by Brittany .
Each diagram in a sequence of 3 diagrams is obtained by drawing a 1 unit square on each side that forms the perimeter of the previous diagram, for example Diagram 2 is obtained by drawing a 1 unit square of the four sides of Diagram 1(on graph paper).Diagram 1 contains one square, diagram 2 four squares and diagram 3 thirteen squares
1)What is the sequence or how many square would diagram 4 contain.
2)What are the appropriate values of that should be inserted at the rows 1, 2 and 3? I just need to know for one and i will find the rest.
We answered this same question just a while ago.
I created the diagrams from your description and counted the 4th diagram to contain 25 squares
BTW, shouldn't the 2nd diagram contain 5 squares ?
So we have the following set of ordered pairs
taking consecute differences of the second column we get 4, 8, 12, ...
(that is, 5-1=4,13-5=8, 25-13=12)
if we take difference of those results we get 4,4,4,...
You can now work backwards and complete the first 2 columns
BTW, since the second difference column consists of a constant value, we know that the original ordered pairs can be represented by a quadratic function.
That way you could calculate the number of squares at any stage, rather than do in by recursion.
Do you know how to find that function?
No i don't know the function and i still don't get how you went from 13 to 25 squares, can you explain how do i come up with that answer for future questions?
I suggest using different colour pencil, simply construct the figures they way you described it
start with 1 square ---> sum = 1
draw a square on all the exposed sides, that would add 4 new squares ---> sum = 5
I then drew a square on each of the exposed sides, adding 8 new ones --->sum = 13
at this point you have 5 rows
top row: 1 square
2nd row: 3 squares
3rd row: 5 squares
4th row: 3 squares
5th row: 1 square ..... total = 13
now if you add another "layer" on this figure you would be adding 8 new squares for a total of 25