prove that ${3^{3n}} + 3^{2n} + 3^{n } + 1$ is divided by $4$. by induction I tried to take the 3 out but it is not helping me much. 
 A: $$...= 3^{2n}(3^n+1)+(3^n+1)= (3^{2n}+1)(3^n+1)$$
Notice that $(3^{2n}+1)$ and $(3^n+1)$ are even, so you are done.
A: You are on the right track. The proposition that you want to prove using induction is $$P(n): 4\mid \underbrace{3^{3n}+3^{2n}+3^n +1}_{:=a_n}$$
It is easy to check that $P(0)$ holds. Assume that $P(n)$ holds for some integer $n\ge 0$ (that is, $a_n$ is a multiple of $4$) and consider $$a_{n+1}=27\cdot 3^{3n}+9\cdot 3^{2n}+3\cdot 3^n+1=3a_n+24\cdot 3^n +6 \cdot3^{2n}-2.$$
Notice that $3a_n$ and $24\cdot 3^n$ are clearly multiples of $4$. Also $$6 \cdot3^{2n}-2 = 2\underbrace{(3^{2n}-1)}_{\text{even}} $$
is a multiple of $4$. Therefore, $a_{n+1}$ is a multiple of $4$. 
A: We have
$$
3^0 \equiv +1 \bmod 4
\\
3^1 \equiv -1 \bmod 4
\\
3^2 \equiv +1 \bmod 4
\\
3^3 \equiv -1 \bmod 4
\\
$$
Therefore, $3^{3n}+3^{2n}+3^{n} + 1 \equiv 1^n +(-1)^n + 1^n + (-1)^n \equiv 0 \bmod 4$.
A: By induction?
Well if we assume $3^{3n} + 3^{2n} + 3^n + 1= 4K$ then 
$3^{3(n+1)} +3^{2(n+1)} + 3^{n+1} + 1=$
$3^{3n+3} + 3^{2n + 2} + 3^{n+1} + 1=$
$3^{3n}*27 + 3^{2n}*9 + 3^n*3 + 1 =$
$[3^{3n} + 3^{2n} + 3^n+1] + 26*3^{3n} + 8*3^{2n} + 2*3^n =$
$4K + 4(6*3^{2n} + 2*3^{2n}) + 2(3^{3n} + 3^n) =$
$4[K+(6*3^{2n} + 2*3^{2n})] + 2*3^{n}(3^{2n} + 1)=$
$4[K+(6*3^{2n} + 2*3^{2n})] + 4*3^{n}*\frac {3^{2n} + 1}2$.
The only thing left to prove is that $3^{2n} + 1$ is even and.... c'mon.....!
A: Let $a_n=3^{3n}+3^{2n}+3^{n} + 1 = (3^3)^n+(3^2)^n+(3^1)^n + (3^0)^n$.
Since 
$$(x-3^3)(x-3^2)(x-3^1)(x-3^0)= x^4 - 40 x^3 + 390 x^2 - 1080 x + 729$$
we have
$$
a_{n+4} =40 a_{n+3} - 390 a_{n+2} + 1080 a_{n+1} - 729a_{n}
$$
The claim that $a_n$ is a multiple of $4$ for all $n$ follows by induction since it is true for $n=0,1,2,3$.
(The actual coefficients in that recurrence do not matter. It only matters that they are integers.)
A: Hint:
Your expression is a geometric series $$\frac{3^{4n}-1}{3-1}$$
You also have that $$3^{4n}=(4-1)^{4n}=\sum_{i=0}^{4n}\binom{4n}{i}4^i(-1)^{4n-i}$$
