Is there a "pattern" in the area when increasing the length and width of a square by 1?

[LordJezo]

1x1 is a 1 area square

2x2 is a 4 area square

3x3 is a 9 area square...


i realize its the square of one of the sides but is there a "pattern" of some sort besides this in the area of the cubes?

 

[dighn]

if side^2 isn't a good enough pattern then i dont know what you are looking for...

 

[Analog]

Originally posted by: LordJezo
1x1 is a 1 area cube

2x2 is a 4 area cube

3x3 is a 9 area cube...


i realize its the square of one of the sides but is there a "pattern" of some sort besides this in the area of the cubes?

Just length times width I guess..

 

[tweakmm]

You're kidding right?

 

[LordJezo]

Originally posted by: tweakmm
You're kidding right?

Nope.

How about.. the area is increased by the length plus the width plus one of the previous figure...

 

[glaHHg]

Originally posted by: LordJezo

Originally posted by: tweakmm
You're kidding right?

Nope.

How about.. the area is increased by the length plus the width plus one of the previous figure...

you just said (a+b)^2 = a^2 + 2ab + b^2 where b = 1.
let b equal anything you want and there's your pattern for computing area when increasing the length by b. area increases from a^2 to (a+b)^2 by 2ab + b^2.

 

[TuxDave]

You mean square right? Square = 2-D, Cube = 3-D

 

[TuxDave]

For this simple case, it all really boils down to manipulating algabra. I could make up another pattern that works, such as:

Notation A(n) = Area with sides n

A(n) = 2*A(n-1) - A(n-2) + 2

(given n > 2)

So for 4x4 = 16, 3x3=9, 2x2=4, it works. 16 = 2*9-4+2

 

[notfred]

Why can't ATers differentiate between area and volume?