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