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Question
Show that \[2^{4n + 4} - 15n - 16\] , where n ∈ \[\mathbb{N}\] is divisible by 225.
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Solution
We have,
\[2^{4n + 4} - 15n - 16 = 2^{4\left( n + 1 \right)} - 15n - 16\]
\[ = {16}^{n + 1} - 15n - 16\]
\[ = \left( 1 + 15 \right)^{n + 1} - 15n - 16\]
\[ =^{n + 1} C_0 {15}^0 +^{n + 1} C_1 {15}^1 +^{n + 1} C_2 {15}^2 + . . . +^{n + 1} C_{n + 1} {15}^{n + 1} - 15n - 16\]
\[ = 1 + (n + 1)15 +^{n + 1} C_2 {15}^2 + . . . +^{n + 1} C_{n + 1} {15}^{n + 1} - 15n - 16\]
\[ = 1 + 15n + 15 +^{n + 1} C_2 {15}^2 + . . . +^{n + 1} C_{n + 1} {15}^{n + 1} - 15n - 16\]
\[ =^{n + 1} C_2 {15}^2 + . . . +^{n + 1} C_{n + 1} {15}^{n + 1} \]
\[ = {15}^2 \left( {}^{n + 1} C_2 + . . . +^{n + 1} C_{n + 1} {15}^{n - 1} \right)\]
\[ = 225\left( {}^{n + 1} C_2 + . . . +^{n + 1} C_{n + 1} {15}^{n - 1} \right)\]
Thus,
\[2^{4n + 4} - 15n - 16\] , where n ∈ \[\mathbb{N}\] is divisible by 225.
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