Question

how to solve this question

Answer 1

0.3+0.33+0.333+……n terms.

Take 3 as common from the given series .

= 3×{0.1+0.11+0.111+…. n terms}

Now we multiply and divide it by 9 .

= 3/9×{0.9+0.99+0.999+…..n terms}

= 3/9×{(1–0.1)+(1–0.01)+(1–0.001)+……n terms}

=> 3/9×{n -(0.1+0.01+0.001+…… n terms) }

Here the term (0.1+0.01+0.001+.....n) becomes a GP

Whose
a=0.1
r= 0.1

Thus sum of the GP series is ...


$s_{n} = \frac{a(1 - {r}^{n} )}{1 - r} \\ s_{n} = \frac{0.1(1 - {0.1}^{n} )}{1 - 0.1} \\ s_{n} = \frac{(1 - 0.1^{n})}{9} \\ hence \: sum \: of \: given \: series \: is \: \\ \frac{3}{9} (n - \frac{(1 - 0.1^{n})}{9}) \\ = \frac{3}{81} (9n - 1 + {0.1}^{n} ) \\ = > \frac{1}{27} (9n - 1 + {0.1}^{n} )$