If we start with a square, cutting each side into fractions, shading one of the corner squares formed by connecting the points of division of opposite sides, and then iterate the process, what protion of the area of the original square is covered by these "corner squares?"

Let's start by defining an iteration.   Iteration is the action or process of repeating an application.

Now let's cut a square into fourths and connecting the points on opposite sides in order to form a smaller square in one corner of our original square.

The bottom yellow square is 1/16 of the area of the original area. We can see this by connecting the points dividing the sides of the original square which shows us that the original square is cut into 16 different smaller squares by these divisions. So, each one of these smaller squares is 1/16 the area of the original square.

We also know that the larger green square formed by in the above figure is 9/16 the area of the original square since, if we were to form all of the divisions of the original square, the green square would contain 9 of these smaller squares that are 1/16 the area of the original square.

Now we will preform the second iteration of this process on the larger of the two squares

We can see that the smaller yellow square is 1/16 the area of the original green square, which means it is (1/16)(9/16) the area of the original square. The blue square is 9/16 the area of the original green square, which means it is (9/16)(9/16) the area of the original square.

Now let's look at the fifth of these iterations.

For this iteration, the small yellow square is 1/16 the area of our fifth square which is (9/16)(9/16)(9/16)(9/16) since it is the fourth division of our original square.  So we must have that the small yellow square is (1/16)(9/16)(9/16)(9/16)(9/16) the area of the original square.

k=i-1
So it would appear from this investigation that the amount of shaded area covered by the "corner squares" is ∑k=0(1/n2)[(n-1)2/n2](k-1) where n is the number of divisions of the square and i is the number of iterations.

Now we will check this formula.  Using GSP we can find the area of each square and see how much of the area of the large square is taken up by the area of these smaller squares.  We can then compare this to the answer we find using our formula.

So we can see that using both methods provide us with the same result, 0.1348.  Let's also look at another case.  For this case we make 8 divisions and will do 6 iterations.

So we can see that using both methods provides us with the same result, 0.0491.  So it appears that this formula does indeed work for this case, and should work for any other case as well, but would this formula also work for other figures?  Let's investigate what happens with a triangle.

How much area of a triangle is covered by an iteration of corner triangles?

Let's start with a triangle ABC and trisect each side, shading the area of the lower left hand triangle.

The large triangle is cut into 9 small triangles of equal area so the yellow triangle is 1/9 the area of the original triangle.

So we must also have that the purple triangle is 4/9 the area of the original triangle since it contains 4 of the smaller 9 triangles from our original triangle.  Now we will divide the purple triangle into smaller triangles.

So by dividing this into smaller triangles we have that the small yellow triangle is 1/9 the area of our purple triangle, therefore it is (1/9)(4/9) the area of the original triangle.  Now we will iterate this process 7 times.

So we should have that each triangle is 1/9 the area of it's larger triangle.  So it would appear that the formula used for the iteration of squares would work for triangles also since there are 3 divisions of each side.  This would mean our n would be 3 and our i would be 7.  Let's use GSP to investigate this hypothesis.

So it appears that the formula for squares works for triangles also.

To see an interactive GSP sketch of the square iteration click here.
To see an interactive GSP sketch of the triangle iteration click here.