The sum of 0.2, 0.22, 0.222, 0.2222.....till n terms is given by

Question

$(a)\;\frac{2}{9}\;[10^{n+1}-10-9n]\qquad(b)\;\frac{2}{81}\;[10^{n+1}+10+9n]\qquad(c)\;\frac{2}{810}\;[10^{n+1}-10-9n]\qquad(d)\;\frac{2}{81}\;[10^{n+1}-10-9n]$

Comments

This is a wrong answer this can also come in chance if u now the than give if u don't now why giving such rubbish answers ok if u don't now then don't give such foolish answer