Use mathematical induction to show that $3^{3n} + 2^{n+2}$ is divisible by 5. Any help would be appreciated thanks!
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2Is it $3^{3n}$ and $2^{n+2}$? As you write the first term, it can be interpreted as $3^3n$. – egreg Sep 29 '13 at 22:59
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Is this homework? – Loki Clock Sep 29 '13 at 23:25
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test preparation – user95126 Sep 30 '13 at 00:30
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it's 3^(3n) thanks – user95126 Sep 30 '13 at 00:30
3 Answers
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If you go all the time with modular arithmetic then
$$3^3=27=2\pmod5\implies 3^{3n}+2^{n+2}=2^n+2^2\cdot 2^n\pmod5=5\cdot 2^n=0\pmod5$$
and we don't actually need direct induction.
DonAntonio
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Well, the induction is in $a\equiv b\mod m\implies a^n\equiv b^n\mod m$ for each $n$, so it is there in disguise. – Pedro Sep 30 '13 at 00:19
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First the base case.
Second: You assume that the statement is true for $n$ and you want to show that it hold for $n+1$. That is, you want to prove that $$3^{3(n+1)}+2^{(n+1)+2}$$ is divisible by $5$.
You are assuming that it holds for $n$, so that means: $5\mid 3^{3n}+2^{n+2}$. In particular, there exists an integer $q$ so that $5q=3^{3n}+2^{n+2}$. Solve for $2^{n+1}$ here and substitute on the first equation. You should be able to factor out a $5$ after you do some algebra.
Daniel Montealegre
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The main observation is that $3^3=27\equiv 2\pmod5$. Using this, the induction step goes like $$3^{3(n+1)}+2^{(n+1)+2} = 3^3\cdot 3^{3n}\, +\, 2\cdot 2^{n+2}= \\ = (25+2)\cdot 3^{3n}+2\cdot 2^{n+2}=\dots $$ Then you can use the induction hypothesis (and that $5\mid 25$).
Berci
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