Question

The sum of n terms of the series 0.3+0.03+0.003+......... is​

Answer 1

3/10+3/100+3/1000....

3(1/10+1/100+1/1000..). a=1/10, r=1/10

3 ×[(1/10(1-(1/10)^n)/(1-1/10)]

3×[1/10{1-(1/10)^n/9/10]

3×{1-(1/10)^n}/9

1-(1/10)^n/3

it will help u

Answer 2

hi sister.....

plz open this attached file.....

hope it helps..