Here it is:
NB: I haven't written every step of working, to save time and effort.
no. of blocks forming one stack = 2n + 1.
total no. of blocks = (2n + 1)^3
Therefore blocks remaining = (2n + 1)(4n^2 + 4n) = f(n)
Basis: for n=1. No. blocks remaining = 3x8 = 24. Therefore divisible by 24.
Assume true for n=k. i.e. (2k + 1)(4k^2 +4k) is divisible by 24.
For n= k + 1.
f( k + 1) = 4(2k +3)(k^2 + 4k + 4)
f(k + 1) - f(k) = 24k^2 + 48k + 48
= 24(k^2 + 2k + 2)
Therefore, f(k + 1) = f(k) + 24(k^2 + 2k + 2)
Both terms of this expression are divisible by 24, therefore f(k + 1) is divisible by 24. It therefore stands by induction that f(n) is divisible by 24 for all integers n>=1.
NB: I haven't written every step of working, to save time and effort.
no. of blocks forming one stack = 2n + 1.
total no. of blocks = (2n + 1)^3
Therefore blocks remaining = (2n + 1)(4n^2 + 4n) = f(n)
Basis: for n=1. No. blocks remaining = 3x8 = 24. Therefore divisible by 24.
Assume true for n=k. i.e. (2k + 1)(4k^2 +4k) is divisible by 24.
For n= k + 1.
f( k + 1) = 4(2k +3)(k^2 + 4k + 4)
f(k + 1) - f(k) = 24k^2 + 48k + 48
= 24(k^2 + 2k + 2)
Therefore, f(k + 1) = f(k) + 24(k^2 + 2k + 2)
Both terms of this expression are divisible by 24, therefore f(k + 1) is divisible by 24. It therefore stands by induction that f(n) is divisible by 24 for all integers n>=1.
If more of us valued food and cheer and song above hoarded gold, it would be a merrier world. - J.R.R Tolkien
