$18884=1\times10^4+80\times111+4$
$288884=2\times10^5+80\times1111+6$
$3888884=3\times10^6+80\times1111+8$
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$100088...8882002=1000\times10^{1003}+80\times\underbrace{1111...11}_\text{1002 tane 1}+2002$
$\sum_{n=1}^{1000}n10^{n+3}+80\times(111+1111+...+\underbrace{1111...11}_\text{1002 tane 1})+\sum_{n=1}^{1000}2n+2$
Bundan sonrasini Mathematica hallediyor..