Is this formula true for $n\geq 1$:$4^n+2 \equiv 0 \mod 6 $? Is this formula true for $n\geq 1$:$$4^n+2 \equiv 0 \mod 6 $$.
Note :I have tried for some values of $n\geq 1$ i think it's true such that
:I used the sum digits of this number:$N=114$,$$1+1+4\equiv 0 \mod 6,1²+1²+4²\equiv 0 \mod 6,1^3+1^3+4^3\equiv 0 \mod 6,\cdots $$ ?
Thank you for any help 

For all $n$,
$$
4^n\equiv1^n\equiv1\pmod{3}
$$
For all $n\gt0$,
$$
4^n\equiv0^n\equiv0\pmod{2}
$$
Therefore, by the Chinese Remainder Theorem
$$
4^n\equiv4\pmod{6}
$$
and therefore, for all $n\gt0,$
$$
4^n+2\equiv0\pmod{6}
$$
