Show that `2^(4n+4) -15n-16`, where `n in N` is divisible by `225.`

Explanation

`2^(4n+4) - 15n - 16 = 16^(n+1) - 15n - 16`

`=(15+1)^(n+1) - 15n - 16`

`=C(n+1,0)15^(n+1) + C(n+1,1)15^(n) +... +C(n+1,n-1)15^2+C(n+1,n)15^1+C(n+1,n+1)15^0 - 15n - 16`

Now, `C(n+1,0)15^(n+1) + C(n+1,1)15^(n) +... +C(n+1,n-1)15^2` is clearly divisible by `225`.

So, we can write it as `225k`.

So, our expression becomes,

`= 225k +C(n+1,n)15^1+C(n+1,n+1)15^0 - 15n - 16`

`=225k+15(n+1)+1-15n-16 `

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