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?

$1+7+7^2+7^3=400$, so $$\begin{array}{rcl}7+7^2+7^3+\cdots+7^{4k}&=&(1+7+7^2+7^3)(7+7^5+7^9+\cdots+7^{4k-3})\\&=&100\cdot 4(7+7^5+7^9+\cdots+7^{4k-3})\end{array}$$
