The fraction 
\[\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},\]
where $n$ is the length of the period of the repeating decimal expansion.  What is the sum $b_0+b_1+\cdots+b_{n-1}$?
$\textbf{(A) }874\qquad \textbf{(B) }883\qquad \textbf{(C) }887\qquad \textbf{(D) }891\qquad \textbf{(E) }892\qquad$
the fraction $\dfrac{1}{99}$ can be written as \[\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}\].
similarly the fraction $\dfrac{1}{99^2}$ can be written as $\sum^{\infty}_{m=1}\dfrac{1}{10^{2m}}\sum^{\infty}_{n=1}\dfrac{1}{10^{2n}}$ which is equivalent to \[\sum^{\infty}_{m=1}\sum^{\infty}_{n=1} \dfrac{1}{10^{2(m+n)}}\]
and we can see that for each $n+m=k$ there are $k-1$ $(n,m)$ combinations so the above sum is equivalent to:
\[\sum^{\infty}_{k=2}\dfrac{k-1}{10^{2k}}\]
we note that the sequence starts repeating at $k = 102$
yet consider \[\sum^{101}_{k=99}\dfrac{k-1}{10^{2k}}=\dfrac{98}{{10^{198}}}+\dfrac{99}{{10^{200}}}+\dfrac{100}{10^{{202}}}=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{100}{10000})=\dfrac{1}{10^{198}}(98+\dfrac{99}{100}+\dfrac{1}{100})=\dfrac{1}{10^{198}}(98+\dfrac{100}{100})=\dfrac{1}{10^{198}}(99)\]
so the decimal will go from 1 to 99 skipping the number 98
and we can easily compute the sum of the digits from 0 to 99 to be \[45\cdot10\cdot2=900\] subtracting the sum of the digits of 98 which is 17 we get
\[900-17=883\textbf{(B) }\qquad\]