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.

Have you ever seen the identity $\frac{q^n-1}{q-1}=1+q^1+q^2+\cdots+q^{n-1}$. It's not hard to prove really it's just polynomial long division. — DRF, Apr 20 '15 at 15:42

score 3 · Accepted Answer · answered Apr 20 '15 at 15:41

3

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

answered Apr 20 '15 at 15:41

mathlove

score 1 · Answer 2 · answered Apr 20 '15 at 15:42

1

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

answered Apr 20 '15 at 15:42

Gregory Grant

score 0 · Answer 3 · answered Apr 20 '15 at 18:30

0

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

answered Apr 20 '15 at 18:30

Bernard

3 Answers3