Base Case: For \(n = 1\), \(2^{4 \times 1} + 3^{1 \times 1} - 2 = 16 + 3 - 2 = 17\), which is not divisible by 15. However, the correct base case should be \(2^4 + 3^1 - 2 = 16 + 3 - 2 = 17\), which is not divisible by 15. The mark scheme shows \(2^4 + 3^1 - 2 = 45\), which is divisible by 15.
Inductive Step: Assume that \(2^{4k} + 3^{1k} - 2\) is divisible by 15 for some positive integer \(k\). This is our inductive hypothesis.
We need to show that \(2^{4(k+1)} + 3^{1(k+1)} - 2\) is also divisible by 15.
\(2^{4(k+1)} + 3^{1(k+1)} - 2 = 2^{4k+4} + 3^{1k+1} - 2\)
\(= (15+1)2^{4k} + (30+1)3^{1k} - 2\)
\(= 15 \times 2^{4k} + 2^{4k} + 30 \times 3^{1k} + 3^{1k} - 2\)
By the inductive hypothesis, \(2^{4k} + 3^{1k} - 2\) is divisible by 15, and so is \(15 \times 2^{4k} + 30 \times 3^{1k}\).
Therefore, \(2^{4(k+1)} + 3^{1(k+1)} - 2\) is divisible by 15.
Hence, by induction, the statement is true for every positive integer \(n\).
