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 Apr 16, 2019
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a)

\(f(n) = 2^{4n+3}+3^{3n+1} \\ f(k+1)-16f(k) = \\ 2^{4(k+1)+3}+3^{3(k+1)+1)}-16\left( 2^{4k+3}+3^{3k+1}\right)\)

 

\(2^{4k+7}+3^{3k+4} - 16\cdot 2^{4k+3}-16\cdot 3^{3k+1}=\\ (16-16)2^{4k+3}+(27-16)3^{3k+1}=11 \cdot 3^{3k+1} =\\ 33 \cdot 3^{3k}\)

 

b)

\(P_1 = f(1) \pmod{11}=0\\ f(1) = 2^3 + 3^1 = 11\\ P_1=True\\ \text{Show }P_n \Rightarrow P_{n+1}\)

 

\(\text{Assume }P_n=True\\ f(n+1) = 2^{4(n+1)+3}+3^{3(n+1)+1} = \\ 16\cdot 2^{4n+3}+27\cdot 3^{3n+1}= \\ 16(2^{4n+3}+3^{3n+1}) + 11\cdot 3^{3n+1} = \\ 16f(n) + 11 \cdot 3^{3n+1} \)

 

\(f(n)\pmod{11} = 0 \text{ by assumption}\\ 11 \cdot 3^{3n+1} \pmod{11} =0 \text{ should be clear}\\ P_n \Rightarrow P_{n+1} \)

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 Apr 16, 2019

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