1/9 {n- (1-(0.1) ⁿ} /9

[0.1+0.11+0.111+………] *9/9

=[0.9+0.99+0.999+…….]*1/9

=[(1-0.1)+ (1-0.01)+ (1-0.001)+….]*1/9

=n-(0.1+0.01+0.001+…….) * 1/9

(0.1+0.01+0.001+…….) is a G.P with common ratio 0.1;

Sum of terms of G.P = a(1-rⁿ)/(1-r)

i.e, (0.1+0.01+0.001+…….) = 0.1(1-0.1ⁿ)/(1-0.1)

i.e,( 1-0.1ⁿ )/9

therefore it becomes , {n-( 1-0.1ⁿ )/9}*1/9

Thanks a lot....