Proof by induction the divisiblity Proof by induction, that 
$$x_n=10^{(3n+2)} + 4(-1)^n\text{ is divisible by 52, when n}\in N  $$
for now I did it like that:
$$\text{for } n=0:$$
$$10^2+4=104$$
$$104/2=52$$

$$\text{Let's assume that:}$$
$$x_n=10^{(3n+2)} + 4(-1)^n=52k$$
$$\text {so else}$$
$$4(-1)^n=52k-10^{3n+2}$$

$$for \text{ n+1}:$$
$$\text {after transforms get something like that:}$$
$$52k=10^{3n+3}$$

But I'm sure, that the last step I did wrong. Actually I don't know when the proof is done, if you would help me I would be thankful.

Assume that for $n\in\Bbb N$, $$10^{(3n+2)} + 4(-1)^n\text{ is divisible by 52.}  $$ This means that $$\frac{10^{(3n+2)} + 4(-1)^n}{52}\in\Bbb Z.$$ Now,
$$\begin{align}
\frac{10^{[3(n+1)+2]} + 4(-1)^{n+1}}{52}&=\frac{10^{(3n+2)}10^3 + 4(-1)^n(-1)}{52}\\
&=\frac{10^{(3n+2)}10^3 + 4(-1)^n(10^3-1001)}{52}\\
&=\frac{10^3\big[10^{(3n+2)}+4(-1)^n\big]+4(-1)^n(-1001)}{52}\\
&=10^3\cdot\frac{10^{(3n+2)} + 4(-1)^n}{52}+\frac{-4004}{52}(-1)^n\\
&=10^3\cdot\frac{10^{(3n+2)} + 4(-1)^n}{52}+(-77)(-1)^n\in\Bbb Z.
\end{align}$$
This implies that $$10^{[3(n+1)+2]} + 4(-1)^{n+1}\quad\text{is divisible by }52.$$
