Just saw this post, and realized that

1/9801 =

0.(000102030405060708091011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192939495969799)(repeat)Similar properties are also exhibited by numbers 998001, 99980001, .. and so on.

Edit: fixed periodic start and end

It is not very obvious to me why this happens. Is there some simple explanation to this property?

**Answer**

There is actually a straightforward reason. As 99 is 1 less than 100, we get a fairly simple expression for its decimal \frac{1}{99}=0.01010101010101\overline{01}\dots Now, \frac{1}{9801}=\left(\frac{1}{99}\right)^2, and the decimal expansion follows from the formula for general power series \left(\sum_{n=1}^\infty x^n\right)^2= x\sum_{n=1}^\infty nx^n.

Letting x=\frac{1}{100}, the decimal expansion for \frac{1}{99} given above is exactly the same thing as writing \frac{1}{99}=\sum_{n=1}^\infty x^n. Applying our identity, the x in front accounts for the double zero. Once n is around 99 we expect to miss a number because we are forcing things to be in decimal, and there will be carrying, which is why the number 98 is missed.

A similar pattern will occur for \frac{1}{998001}=\left(\frac{1}{999}\right)^2, since as before \frac{1}{999}=0.001001\overline{001}.

**Attribution***Source : Link , Question Author : Lazer , Answer Author : Eric Naslund*