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Prove that $3^{n+1}+3^n+3^{n-1}$ is divisible by $13$ for all positive integral values of $n$.

I tried: $3^n \cdot 3^1+3^n+3^n\cdot\frac{1}{3}$

Then what should I do next? Help please?

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    $\begingroup$ Start again: common factor $3^{n-1}$. In divisibility questions, "fractions" can lead to confusion. $\endgroup$ – André Nicolas Feb 12 '14 at 18:47
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I'm not sure how I would proceed from your path, but here is an alternative: $$3^{n+1} + 3^n + 3^{n-1}= 3^{n-1}(3^2+3+1)=3^{n-1} \cdot 13$$ So: $$\frac{3^{n+1} + 3^n + 3^{n-1}}{13}=3^{n-1}$$ And since $n \in \mathbb{N}$: $$13|(3^{n+1} + 3^n + 3^{n-1})$$

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It can be finished your way: $\ 3^n(3+1+\overbrace{1/3}^{\large \color{#c00}{-4}}) \,\equiv\, 0,\ $ by $\ \dfrac{1}3\equiv \dfrac{-12}{\ 3}\equiv \color{#c00}{-4}\pmod{13}$

But here there is no need to use fractions, since we can proceed fraction-free as in Dror's answer. Furthermore, one needs to be very careful with fractions in modular arithmetic lest one do the equivalent of dividing by zero (or a zero-divisor). To avoid that error we must ensure that we use only fractions with denominator coprime to the modulus.

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