From the given series, take common factor 3
Thus
S=3(1+11+111+1111+...............nthterm)
Or
S=3{1+(1+10)+(1+10+100)+(1+10+100+1000)+...tonterms}
Thus the nth term.
tn=1+10+100+..............+10n−1
This is a G.P. where first term is 1 and common ratio is 10 ,
Thus sum of n terms is
tn=1(1−10n)1−10tn=(10n−1)9
Or
Thus
S=3∑ni=1tn=3∑ni=110n−19
Hence the sum,
S=3∑10n9−3∑19
Or
S=3{10(1−10n)(1−10).9−(19).n}
Or
S=3{10.(10n−1)81−n9}
Thus,
S={10.(10n−1)27−n3}=10n+1−10−9n27