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?
 A: $1+7+7^2+7^3=400$, so $$\begin{array}{rcl}7+7^2+7^3+\cdots+7^{4k}&=&(1+7+7^2+7^3)(7+7^5+7^9+\cdots+7^{4k-3})\\&=&100\cdot 4(7+7^5+7^9+\cdots+7^{4k-3})\end{array}$$
A: 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 $n$.
A: Possibly easier:
$$\eqalign{7+7^2+\cdots+7^{4n}
  &=(7+7^2+7^3+7^4)(1+7^4+\cdots+7^{4n-4})\cr
  &=2800(1+7^4+\cdots+7^{4n-4})\ .\cr}$$
A: Hint
You have to show $7+7^2+7^3+...+7^{4n}\equiv 0\mod 100$.
Now observe $7^4=49^2=2401\equiv 1\mod 100$, and, as you showed, the sum is
$$7+7^2+7^3+...+7^{4n}=\frac{ 7(7^{4n}-1)}6.$$
A: 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
$$
\begin{align}
&\sum_{k=0}^n\left(7^{4k+1}+7^{4k+2}+7^{4k+3}+7^{4k+4}\right)\\
&=\underbrace{\sum_{k=0}^{n-1}\left(7^{4k+1}+7^{4k+2}+7^{4k+3}+7^{4k+4}\right)}_{0\bmod{400}\text{ by inductive hypothesis}}
+7^{4n}\underbrace{\vphantom{\sum_{k=0}^{n-1}}\left(7^1+7^2+7^3+7^4\right)}_{0\bmod{400}}\\[6pt]
&\equiv0\pmod{400}
\end{align}
$$
A: 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 $n+1$:
$$\begin{align}\sum_{i=1}^{4(n+1)}7^i=&\sum_{i=1}^{4n}7^i+7^{4n+1}+7^{4n+2}+7^{4n+3}+7^{4n+4} =\\ &100a+7^{4n+1}+7^{4n+2}+7^{4n+3}+7^{4n+4}
=\\ &100a+7^{4n}(7+7^2+7^3+7^4)=\\ &100a+7^{4n}\cdot 2800=\\ & (a+7^{4n}\cdot 28)\cdot 100= \\ &100\cdot b\end{align}$$
