How would I find the decimal expansion of $1/99^2$

Question

I want to find the repeating decimal expansion of $1/99^2$. All I know is that $1/99 = 0.010101\cdots$. How would I continue?

Answer 1

Partial answer: it's enough find the decimal places for $$ \frac{10000}{99^2}=\frac{1}{0.99^2}=\frac{1}{(1-0.01)^2}. $$ Recall that $\frac{1}{(1-x)^2}\approx 1+2x+3x^2+4x^3+5x^4+6x^5+\cdots$, we have (for $x=0.01$) $$ \frac{10000}{99^2}=1.0203040506070809... $$ which works out nicely until you get to terms like $100x^{99}$ and above. In any case, this should give you about the first $200$ decimal places for $\frac{1}{99^2}$.

Answer 2

Note that for all $x \in (-1,1)$, we have $$ \frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots $$ Now, let $x = .01$

Answer 3

According to Mathematica, $1/99^2$ is

\begin{array}{l} 0. \overline{00010203040506070809101112131415161718192021222324252627282930313} \\ \overline{2333435363738394041424344454647484950515253545556575859606162636465} \\ \overline{666768697071727374757677787980818283848586878889909192939495969799} \end{array}