Question

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

Answer 1

We have `2^(4n + 4) - 15n - 16`

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

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

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

= `""^(n + 1)"C"_0  15^0 + ""^(n + 1)"C"_1  15^1 + ""^(n + 1)"C"_2  15^2 + ""^(n + 1)"C"_3  15^3 + ... + ""^(n + 1)"C"_(n + 1) (15)^(n + 1) - 15n - 16`

= `1 + (n + 1)15 + ""^(n + 1)"C"_2  15^2 + ""^(n + 1)"C"_3  15^3 + ... + ""^(n + 1)"C"_(n + 1) (15)^(n + 1) - 15n - 16`

= `1 + 15n + 15 + ""^(n + 1)"C"_2  15^2 + ""^(n + 1)"C"_3  15^3 + ... + ""^(n + 1)"C"_(n + 1)  (15)^(n + 1) - 15n - 16`

= `15^2 [""^(n + 1)"C"_2 + ""^(n + 1)"C"_3  15 + ... "so  on"]`

Thus, `2^(4n + 4) - 15n - 16` is divisible by 225.