# nth term investigation

• October 19th 2005, 11:29 AM
rpatel
nth term investigation
hello.

i have been given a task to investigate a series of 3d shapes. i have attached a table below show the amount of border squares and darkers squares.

i have also attached within the document what the first 3D shape looks like. but really i am after the nth term i have been told to use forumla : a n^3 + b n ^2 + cn + d

please can someone help. i am sturggling so much.

i need 3 sets of equations. for total number of border squares, darker squares and total squares.

thanks
• October 19th 2005, 05:22 PM
hpe
Quote:

Originally Posted by rpatel
hello.

i have been given a task to investigate a series of 3d shapes. i have attached a table below show the amount of border squares and darkers squares.

i have also attached within the document what the first 3D shape looks like. but really i am after the nth term i have been told to use forumla : a n^3 + b n ^2 + cn + d

Here's how I understand your problem:
The 1x1 case is a cube, with 6 cubes stuck on the 6 faces. Result: 1 "dark" cube, 6 cubes that you can see, total of 7 cubes.
The 3x3 case is the object from the 1x1 case, with 18 cubes stuck on the faces of the 7 cubes. Result: 7 "dark" cubes (that's the 7 original from the 1x1 object), 18 cubes that you can see, total of 25 cubes.
The 5x5 case is the object from the 3x3 case, with 38 cubes stuck on the faces of the 25 cubes. Result: 25 "dark" cubes (that's the 25 from the 3x3 object), 38 cubes that you can see, total of 63 cubes.
And so on.
Now to fit the number of "dark" cubes to a formula $an^3 + bn^2 +cn + d$, where n = 1, 2, 3, 4, you get the equations
$a + b + c + d = 1$
$8a + 4b + 2b + d = 7$
$27a + 9b + 3c + d = 25$
$64a + 16b + 4c + d = 63$.
The solution is $a=\frac{4}{3}, \, b = -2, \, c = \frac{8}{3}, \, d = -1$.
So the number of dark cubes for the n-th object is $\frac{4}{3}n^3 -2n^2 + \frac{8}{3}n -1$. Believe it or not, this is always an integer.

The number of total cubes for the n-th object is the same as the number of dark squares for the (n+1)th object, hence $\frac{4}{3}(n+1)^3 -2(n+1)^2 + \frac{8}{3}(n+1) -1$ - you can expand this if you want.

Finally the number of visible cubes for the n-th object is the difference of these two expressions.

Hope this helps. :) btw, this is definitely not grade school stuff.
• October 20th 2005, 07:10 AM
rpatel
thank you so so much

:D

regards