<s>[INST] Patterns in $\frac{80}{81}$ and $\frac{10}{81}$. The decimal form of $\frac{80}{81}$ is $0.987654320\ldots$ notice the expected $1$ is missing. The decimal form of $\frac{10}{81}$ is $0.12345679\ldots$ notice the expected $8$ is missing. Can someone expansion why the decimal form is the way they are? I think it has something to do with $(10-1)^2=81$.
 [/INST] This is because
$\dfrac1{(1-x)^2}
=\sum_{n=0}^{\infty} (n+1)x^n
$.
Putting $x=.1$,
this is
$\dfrac1{.9^2}
=\dfrac1{.81}
=\dfrac{100}{81}
=1+2/10+3/100+4/1000 + ...
$
so
$\dfrac{10}{81}
=1/10+2/100+3/1000+4/10000 + ...
=0.1234567...
$.
The next terms are
$8/10^8+9/10^9 +10/10^{10}+...
$,
but we get a carry here
(from the $10/10^{10}$)
and these terms have a value of
$9/10^8+0/10^9 +0/10^{10}+...
$
which explains the
$....6790...$.
The other is just $1-x$
where $x$ is a decimal:
$\dfrac{80}{81}
=1-\dfrac{1}{81}
$.
</s>