repeating unit of 1's 
Tried getting the last 10 digits calculated out, but couldn't figure out a pattern for the rest of more than 90 digits. Would appreciate any clue. Thanks!
 A: A deleted answer had the right idea but just missed.
Start with just 1.  Now add in the next nine repunits, from 11 to 1,111,111,111.  The ones place adds up to 9, which must be added to the previous 1 to give 10 and thus carry a 1 over to the tens place.  The preceding places give 123,456,789, which together with the carried 1 gives 123,456,790.  So the sum of the first ten repunits is 1,234,567,900.
Now put in nine more repunits.  The ten place values from the right add up to 9,999,999,999, which together with the previous 1,234,567,900 gives 1,234,567,899 and a carry digit of 1.  The nine "new" place values again give 123 456,790 including the carried unit, so the sum of the first 19 repunits will be
1,234,567,901,234,567,899.
Keep adding in groups of nine repunits until you have a hundred of them.  You should find a sum with the form
1,234,567,901,234,567,901,...901,234,567,890
where the block 123456790 is repeated ten times (once for each group of nine repunits after the first ten) before the final ten digits.  Sum the digits from there.
A: Note that  $$1=10^0\\
11 = 10^1+10^0\\
111 = 10^2+10^1+10^0\\
1111 = 10^3+10^2+10^1+10^0\\ 
\vdots\\
111...111 (\text{hundred}~ 1's) = 10^{99}+10^{98} \cdots +10^1+10^0$$
So, adding all the above equations, the sum of first $100$ repunits will be given as:
$$ S = 1+11+\cdots + 111...111 = 1 \cdot 10^{99}+2\cdot10^{98}+\cdots+ i\cdot10^{100-i}+\cdots+100\cdot10^0$$
Now, I leave out to you to find out the sum of the digits of $S.$
