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?

It remains to show that $7^{4n} - 1$ is a multiple of $600$.

Since $600 = (2^3)(3)(5^2)$, the goal is equivalent to showing that the three congruences
\begin{align*}
7^{4n} &\equiv 1\;(\text{mod}\;2^3)\\[4pt]
7^{4n} &\equiv 1\;(\text{mod}\;3)\\[4pt]
7^{4n} &\equiv 1\;(\text{mod}\;5^2)\\[4pt]
\end{align*}
hold for all positive integers $n$.

Now simply note that
\begin{align*}
7 &\equiv -1\;(\text{mod}\;8)\\[4pt]
7 &\equiv 1\;(\text{mod}\;3)\\[4pt]
7^2 &\equiv -1\;(\text{mod}\;25)\\[4pt]
\end{align*}
Can you finish it?

For an inductive approach, note that
\begin{align*}
S_{4(n+1)}-S_{4n}
&=
\frac{7(7^{4(n+1)}-1)}{6}-\frac{7(7^{4n}-1)}{6}
\\[4pt]
&=
\frac{7(7^{4(n+1)}-7^{4n})}{6}
\\[4pt]
&=
\frac{7^{4n+1}(7^4-1)}{6}
\\[4pt]
&=
\frac{7^{4n+1}(2400)}{6}
\\[4pt]
&=7^{4n+1}(400)
\\[4pt]
\end{align*}
hence, if $S_{4n}$ is a multiple of $100$, then so is $S_{4(n+1)}$.

Since $S_{4n}$ is a multiple of $100$ when $n=1$, it follows (by induction on $n$), that $S_{4n}$ is a multiple of $100$, for all positive integers $n$.
