April 13th, 2019, 02:29 AM  #1 
Member Joined: Oct 2012 Posts: 77 Thanks: 0  the sum
Find the sum of 1+11+111+1111+ .....1.....1 to the n th term in term of n. 
April 13th, 2019, 05:53 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 
Enter "1+11 + 111+1111" here: The OnLine Encyclopedia of Integer SequencesÂ® (OEISÂ®) 
April 13th, 2019, 05:56 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,751 Thanks: 2135 
$(10^{n+1}  9n  10)/81$

April 13th, 2019, 07:02 AM  #4 
Member Joined: Oct 2012 Posts: 77 Thanks: 0 
1+11+111+1111+11111+..... 1/9(9+99+999+9999+99999+......) =1/9(101+10^21+10^31+.......) =1/9(10(10^n1)/9n)=(10^(n+1)9n10)/81 