How is $\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$?

Question

As the title states, how is: $$\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$$

I can't see the pattern. Can someone please help? Thanks.

Answer 1

Note that for $x\not=1$, $$(x-1)(x^5+x^4+x^3+x^2+x+1)=x^6-1$$ $$\Rightarrow \frac{x^6-1}{x-1}=x^5+x^4+x^3+x^2+x+1$$

Answer 2

It is true in general that $x^6-1=(x-1)(1+x+x^2+\cdots+x^5)$. Just substitute $x=10^4$.

Answer 3

It all comes from the algebraic identity: $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\dots+1).$$