Parts of a cuboid after n cuts

[gjorgensen]

Homework Statement

I have been asked to find a recursive formula for the number of parts of a cuboid after n cuts, and then prove my formula.

The Attempt at a Solution

I have through a 3D drawing program figured out a the number of parts after 8 cuts
cuts, parts
1,2
2,4
3,8
4,15
5,25
6,38
7,54
8,73

I have noticed a relationship for n >2, which is that R(n+1)=2R(n) - (R(n-1)+3)
However I am looking for a recursive formula which will work for all n, and I have no idea how to prove this...

I would appreciate all the help I can get!

In advance, thank you!

 

[tiny-tim]

welcome to pf!

hi gjorgensen! welcome to pf!

(try using the X₂ icon just above the Reply box )

R(n+1)=2R(n) - (R(n-1)+3)

However I am looking for a recursive formula which will work for all n …

(that is a recursive formula … you mean you want a non-recursive one! )

As with any https://www.physicsforums.com/library.php?do=view_item&itemid=158" , first find the general solutions to the "complementary equation", in this case R_n+1 - 2R_n + R_n-1 = 0,

and then add any particular solution to the complete equation (guess a polynomial)

 

[gjorgensen]

hi gjorgensen! welcome to pf!

(try using the X₂ icon just above the Reply box )


(that is a recursive formula … you mean you want a non-recursive one! )

As with any https://www.physicsforums.com/library.php?do=view_item&itemid=158" , first find the general solutions to the "complementary equation", in this case R_n+1 - 2R_n + R_n-1 = 0,

and then add any particular solution to the complete equation (guess a polynomial)

Thanks a lot, but what I meant was that the recursive formula I stated does not work for n=0,1,2. And I know that there exists a recursive formula which works for all n. Any suggestions, or am I misunderstanding something?

again, thank you!