Sprint Chief

Thanks speedking and for the tidied-up version!

Tough question - is that really GCSE level? Needs a bit of lateral thinking.

My process was to use 3D triangle numbers - view the whole thing as a "cube" of blocks and one corner is made up of a 3D triangle number

= n (n+1) (n+2) / 6

If I double this I get one "edge" of the cube - but I've replicated the cubes in the middle, which make up a 2D triangle number, so I have to subtract the 2D triangle number

= 2 n (n+1) (n+2) / 6 - n (n+1) / 2

Double this complete expression and you get one half of the cube, but this time we've double counted a bunch more cubes - if I subtract two 2D triangle numbers I've then double subtracted the centre bunch so they have to be added back in!

= 2 * [ 2 n (n+1) (n+2) / 6 - n (n+1) / 2 ] - 2 n (n+1) / 2 + n

Then finally double the two halves of the cube to give the whole lot - but this time we've double counted four 2D triangle numbers, less 4 plus the repeated centre block!

= 2 * { 2 * [ 2 n (n+1) (n+2) / 6 - n (n+1) / 2 ] - 2 n (n+1) / 2 + n } - 4 n (n+1) / 2 + 4 n - 1

Multiply all this lot out and you should get something like the above expression (typos excepted ), as simplified by speedking. And if you can do that under exam conditions you deserve the GCSE and then some!

fast bloke

hang on - its a GCSE question

"the formula must be some way of describing the total number of cubes in relation to the black and white cubes "

total=black+white

Ted Maul

doing your sons homework?? tut tut