can you some how manipulate it in geometric progression .

\(\frac{1}{9}(9+99+999+9999………)\)

now can u transfer it into G.P series

ok let’s see

\(\frac{1}{9}[(10^1-1)+(10^2-1)+………] \) upto 2k+1 terms

= \(\frac{1}{9}[(10+10^2+10^3+……)-(1+1+1+1…….)]\)

now can u see the G.P Series , with first term 10 and common ratio 10

use the formula ,\( \frac{a(1-r^n)}{r-1}\)

so the final ans is \(\frac{1}{9}[\frac{10}{9}(10^{2k+1}-1)-n]\)