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## Why does the number $$98$$ disappear when writing the decimal expansion of $$\frac{1}{9801}$$ ?

### Solution

We have $\frac{1}{99 \cdot 99}=\frac{1}{9801}$ $=0.000102030405 \ldots 9596979900010203 \ldots$ Now, let’s find the MacLaurin / Taylor Series for $$f(x)=\frac{1}{1-x}$$
$\frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}+\cdots\;\;\;\;\;\;\;|x|<1$ $\Rightarrow\frac{1}{(1-x)(1-x)}=1+2 x+3 x^{2}+4 x^{3}+5 x^{4}+\cdots\;\;\;\;\;\;\;|x|<1$
Let $$x=10^{-2}$$, we can explain the pattern expanded from $$\frac{1}{9801}$$. the reason for disappearance of the number $$98$$ is caused by the propagation of the carry $$1$$ from $$100x^{99}$$. Similar patterns can be deduced by putting $$x=10^{-1}, 10^{-3}, 10^{-4}, 10^{-5}, \ldots$$ to get $\frac{1}{9 \cdot 9}, \frac{1}{999 \cdot 999}, \frac{1}{9999 \cdot 9999}, \frac{1}{99999 \cdot 99999}, \ldots$
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Why does the number $$98$$ disappear when writing the decimal expansion of $$\frac{1}{9801}$$ ?
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