How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is natural number)?

Question

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.

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

Answer 1

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 $n$.

Answer 2

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

Answer 3

Possibly easier: $$\eqalign{7+7^2+\cdots+7^{4n} &=(7+7^2+7^3+7^4)(1+7^4+\cdots+7^{4n-4})\cr &=2800(1+7^4+\cdots+7^{4n-4})\ .\cr}$$

Answer 4

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 $$7+7^2+7^3+...+7^{4n}=\frac{ 7(7^{4n}-1)}6.$$

Answer 5

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 $$ \frac{7\left(7^{4n}-1\right)}{6}\equiv0\pmod{400} $$

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

Answer 6

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 $n+1$: $$\begin{align}\sum_{i=1}^{4(n+1)}7^i=&\sum_{i=1}^{4n}7^i+7^{4n+1}+7^{4n+2}+7^{4n+3}+7^{4n+4} =\\ &100a+7^{4n+1}+7^{4n+2}+7^{4n+3}+7^{4n+4} =\\ &100a+7^{4n}(7+7^2+7^3+7^4)=\\ &100a+7^{4n}\cdot 2800=\\ & (a+7^{4n}\cdot 28)\cdot 100= \\ &100\cdot b\end{align}$$