Respuesta :
Answer:
9(1/10) [1+(1/10)+(1/10)^2+(1/10)^3+.....]
Step-by-step explanation:
0.99999...= 9(1/10) + 9(1/10)^2+9(1/10)^3+9(1/10)^4+.....
taking 9(1/10) common,
=9(1/10) [1+(1/10)+(1/10)^2+(1/10)^3+.....]
therefore, this is the required geometric series.
- the sum of the infinite series: 1+r+r^2+r^3+r^4+r^5+r^6....
is, [tex]\frac{1}{1-r}[/tex]
- the series inside the square brackets are infinite geometric progression series.
so the whole sum = 9(1/10)[[tex]\frac{1}{1-\frac{1}{10} }[/tex]]
=9(1/10)[10/9]
=1
