How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is natural number)? How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is entire number) ?  
I thought to calculate $S_{4n}$ according to:
$$ S_{4n} = \frac{7(7^{4n}-1)}{7-1} = \frac{7(7^{4n}-1)}{6} $$  
But know, I don't know how to continue for get what that rquired.  
I will be happy for help or hint.  

After beautiful ideas for solving this question, someone know how to do it with induction too?

Using the Formula for a Geometric Series
We need to show more than $7^4\equiv1\pmod{100}$. If that were all we knew, then because $2\mid6$, all we would know would be
$$
\frac{7\left(7^{4n}-1\right)}{6}\equiv0\pmod{50}
$$
However, since $7^4=2401\equiv1\pmod{800}$, we know that
$$
\frac{7\left(7^{4n}-1\right)}{6}\equiv0\pmod{400}
$$

Using Induction
Note that
$$
7^1+7^2+7^3+7^4=2800\equiv0\pmod{400}
$$
Therefore, if
$$
\sum_{k=0}^{n-1}\left(7^{4k+1}+7^{4k+2}+7^{4k+3}+7^{4k+4}\right)\equiv0\pmod{400}
$$
then
$$
\begin{align}
&\sum_{k=0}^n\left(7^{4k+1}+7^{4k+2}+7^{4k+3}+7^{4k+4}\right)\\
&=\underbrace{\sum_{k=0}^{n-1}\left(7^{4k+1}+7^{4k+2}+7^{4k+3}+7^{4k+4}\right)}_{0\bmod{400}\text{ by inductive hypothesis}}
+7^{4n}\underbrace{\vphantom{\sum_{k=0}^{n-1}}\left(7^1+7^2+7^3+7^4\right)}_{0\bmod{400}}\\[6pt]
&\equiv0\pmod{400}
\end{align}
$$
