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

Explanation

We have
`2^(4n+4)-15n - 16 = 2^(4(n+1))-15n - 16 = 16^((n+1))- 15n -16.`
Now, `16^(n+1)-15n-16=(1+15)^(n+1)-15n -16`
`=^(n+1) C_(0)+^(n+1) C_(1) xx 15 + ^(n+1)C_(2) xx (15)^(2) + ^(n+1)C_(3) xx(15)^(3)+...+.^(n+1)C_(n+1) xx(15)^(n+1)-15n-16`
`= 1+(n+1) xx 15 + .^(n+1)C_(2) xx(15)^(2)+ ^(n+1) C_(3) xx(15)^(3)+...+ (15)^(n+1)-15n-16`
`=.^(n+1)C_(2) xx (15)^(2) = .^(n+1)C_(3) xx (15)^(3) +...+(15)^(n+1)`
`(15)^(2) xx[.^(n+1)C_(2) + .^(n+1)C_(3) xx 15 +...+(15)^(n-1) ]`
`=225 xx` (an integar).
Hence, `(2^(4n+4)-15n-16)` is divisible by 225.