Prove that $13\vert(3^{n+1} +3^{n} +3^{n-1})$

Question

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

Answer 1

$3^{n+1}+3^n+3^{n-1}=3^{n-1}(**+**+**)$

Can you see it now???

Answer 2

Besides the obvious factorization approach, you can reason by induction.

$$3^{(n+1)+1}+3^{(n+1)}+3^{(n+1-1)}=3\cdot 3^{n+1}+3\cdot 3^n+3\cdot 3^{n-1}=3 \cdot (3^{n+1}+3^n+3^{n-1}),$$ and $$3^{1+1}+3^{1}+3^{1-1}=13.$$ Hence $$S_{n+1}=3\cdot S_n\text{, and }S_1=13.$$

Answer 3

Observe that $$1\bmod13=1,\\3\bmod13=3,\\9\bmod13=9,\\27\bmod13=1,\\81\bmod13=3,\\243\bmod13=9,\\...$$ The modulos repeat periodically and the sum on a period is $13$.