Can I prove this using modular congruence? $3 \mid 5^n + 2\cdot11^n$ [EDIT] I'd prove this by induction, and I'd been thinking in how prove that using congruence. Please help.

[EDIT] 
I can got to in (−1)^n+(−1)^n+1 mod 3, but I don't know if this it's a congruent mod 3 AAAAAAAAA
If you look at the remainders when each number in your expression is divided by $3$ then,

$5^n + 2\cdot11^n \equiv (-1)^n + (-1)\cdot (-1)^n \equiv (-1)^n + (-1)^{n+1} \equiv 0 \mod 3$

This is because if $n$ is odd, $(-1)^n = -1$ and $(-1)^{n+1} = 1$ hence their sum is zero which is divisible by three. Similarly if $n$ is even, $(-1)^n = 1$ and $(-1)^{n+1} = -1$ and their sum is again zero which is divisible by $3$.
1) $5^n + 2*11^n \equiv 2^n+2*2^n \pmod 3\equiv$

$2^n + 2^{n+1}\equiv (-1)^n+(-1)^{n+1}\equiv$

$(-1)^n - (-1)^n \equiv 0\pmod 3$.

So $3|5^n + 2*11^n\pmod 3$.

2) induction

If $n= 1$ then $5^1 + 2*11 = 27$ and $3|27$.

If we assume $3|5^k + 2*11^k$ then
Then

$5^{k+1} +2*11^{k+1} =$

$5^k*5 + 2*11^k*11 =$

$5(5^k+ 2*11^n) + 6*11^k$

So as $3|5^k + 11^n$ and $3|6*11^k$ we have $3|5(5^n+2*11^k) + 6*11^k$.

3) Fermats little Theorem.

$a^2 \equiv 1 \pmod 3$ for any $a$ relatively prime to $3$.

So if $n=2m$ is even then

$5^{2m} + 2*11^{2m}\equiv (5^2)^m+2*(11^2)^m \equiv 1+2*1 \equiv 0 \pmod 3$.

And if $n = 2m+1$ is odd then 

$5^{2m+1} + 2*11^{2m+1}\equiv 5+ 2*11\equiv 27 \equiv 0 \pmod 3$.