Answer to Question #127808 in Discrete Mathematics for Mullah Dobgangha Tony

Question #127808
Problem B
Show that 3 · 4n + 51 is divisible by 3 and 9 for all positive integers n.
1
Expert's answer
2020-07-29T12:50:17-0400

For any integer n ≥ 1, let PnP_n be the statement that 34n+513\cdot4^n+51 is divisible by 9.


Base case. The statement P1P_1 says that 341+51=633\cdot4^1+51=63 is divisible by 9, which is true.

Inductive Step. Fix k ≥ 1, and suppose that PkP_k holds, that is, 34k+513\cdot4^k+51 is divisible by 9,


34k+51=9m,mZ,m73\cdot4^k+51=9m, m\in \Z, m\geq7

It remains to show that Pk+1P_{k+1} holds, that is, 34k+1+513\cdot4^{k+1}+51 is divisible by 9,


34k+1+51=4(34k)+51=3\cdot4^{k+1}+51=4\cdot(3\cdot4^k)+51==4(34k+51)451+51==4\cdot(3\cdot4^k+51)-4\cdot51+51=

=49m351=9(4m17),mZ,m7=4\cdot9m-3\cdot51=9(4m-17),m\in \Z, m\geq7

Therefore Pk+1P_{k+1} holds.

Thus, by the principle of mathematical induction, for all n ≥ 1, PnP_n holds.

Therefore 34n+513\cdot4^n+51 is divisible by 9 for all positive integers n.

Therefore 34n+513\cdot4^n+51 is divisible by 3 for all positive integers n.



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