t0rajir0u 14.05.2006 08:35 Still a little easy for Pre-Olympiad... Write the $n$-digit repdigit $111...$ as $\frac{10^n - 1}{9}$. $\sum_{k=1}^{n} \frac{10^n - 1}{9} = \frac{1}{9} \left( \sum_{k=1}^{n} 10^n \right) - \frac{n}{9} = \boxed{ \frac{10^{n+1} - 10 - 9n}{81} }$
Tanjiro139 27.02.2022 05:22 Let $A=1+11+111+\cdots+\underbrace{111\ldots111}_{n\text{ digits}}$ this expression can be written as $\sum_{i=0}^{n-1}(n-i)\cdot10^i = \frac{10^{n+1} -9n-10 }{81}$