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?

Induction proof of 
$$\sum_{i=1}^{4n}7^i=100a.$$
For $n=1$, it is true: 
$$7^1+7^2+7^3+7^4=2800=100\cdot 28.$$
Assuming it is true for $n$:, prove for $n+1$:
$$\begin{align}\sum_{i=1}^{4(n+1)}7^i=&\sum_{i=1}^{4n}7^i+7^{4n+1}+7^{4n+2}+7^{4n+3}+7^{4n+4} =\\ &100a+7^{4n+1}+7^{4n+2}+7^{4n+3}+7^{4n+4}
=\\ &100a+7^{4n}(7+7^2+7^3+7^4)=\\ &100a+7^{4n}\cdot 2800=\\ & (a+7^{4n}\cdot 28)\cdot 100= \\ &100\cdot b\end{align}$$
