- #1

Tsunoyukami

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- 11

*"Prove ##5|(3^{3n+1}+2^{n+1})## for every positive integer ##n##."*(Exercise 11.8 from

*Mathematical Proofs: A Transition to Advanced Mathematics*3rd edition by Chartrand, Polimeni & Zhang; pg. 282)

I'm having difficulty solving this exercise. My first thought was to use induction but I get stuck with an expression that I can't seem to manipulate into the required form. The next thing I tried was considering cases (when ##n## is even and when ##n## is odd) but once again I get stuck.

Can anyone suggest another method of proof?

I feel that induction should work out for me despite my difficulties. As such, I will explain what I've done so far. If you would like to see work I describe but do not include below let me know and I'll include it.

First I showed that this expression holds for the case ##n=1##. I assumed this holds for some arbitrary positive integer ##k##. Now I wish to show that ##5|(3^{3(k+1)+1}+2^{(k+1)+1})##.

From here I tried working with the expression ##3^{3(k+1)+1}+2^{(k+1)+1}##:

##3^{3(k+1)+1}+2^{(k+1)+1}=3^{3k+3+1}+2^{k+1+1}=3^{3k+4}+2^{k+2}=3^{3k} \cdot 3^4 + 2^{k} \cdot 2^{2}=81\cdot3^{3k}+4\cdot2^{k}##

This is where I get stuck - I don't know how to massage this equation to show that 5 divides it.