I was asked to prove, using induction, that 3(adsbygoogle = window.adsbygoogle || []).push({}); ^{4n+2}+ 2^{6n+3}is divisible by 17. I tried to do it, but I couldn't get anywhere. Can someone give me a push in the right direction?

Here's my attempt:

f(k) = 3^{4k+2}+ 2^{6k+3}

f(k+1) = 3^{4k+6}+ 2^{6k+9}

And now, all I have to do is prove that f(k+1) - f(k) = 17m, but I couldn't do it.

I don't really see why induction is necessary anyway. Here's my induction-free attempt:

[tex]3^{4n+2} + 2^{6n+3} = 9^{2n+1} + 8^{2n+1} = (9+8) \sum^{2k+1}_{n=1} 9^{2k+1-n} \; 8^{n-1} = 17m[/tex]

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# Prove, using induction

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