Prove that f(n) = 4^n + 6n - 1 is divisible by 9 for all positive integers of n.(adsbygoogle = window.adsbygoogle || []).push({});

I've proved this by considering that f(k) is divisible by 9, i.e f(k) = 4^k + 6k -1 = 9m where m is some integer. Rearranging to give 4^k = 9m - 4k + 1 and then considering f(k+1) = 4^(k+1) + 6(k+1) - 1 then substituting 4^k and showing it is divisible by 9, which works.

I've been trying to do it another way i.e by considering f(k+1) - f(k), doing so I get

[itex] f(k+1) -f(k) = 4^{k+1} + 6k + 6 -1 -4^k -6k + 1 [/itex]

and then going on to get [itex] f(k+1) -f(k)= 3(4^k) + 6 [/itex]

however i'm stuck here, I can obviously see that this is divisible by 9, but how can I show it?

EDIT: Similarly how do I show that 120(2^(4k)) + 78(3^(3k)) is divisible by 11?

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# Homework Help: Mathematical induction

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