The sum of the first 10 terms of the series 9+99+999+ . . . Is

Options

A. `(9)/(8)(9^(10)-1)`
B. `(100)/(9)(10^(9)-1)`
C. `10^(9)-1`
D. `(100)/(9)(10^(10)-1)`

Answer

B

Explanation

Let,
`S_(n)=9+99+999+ . . .. n` terms
`rArrS_(n)=(10-1)+(100-1)+(1000-1_+ . . .n` terms
`rArrS_(n)=10+10^(2)+10^(3))+ . . .n` terms)-(1+1_ . ..n terms)
`rArrS_(n)=(10(10^(n)-1))/(10-1)-n`
`[becausea+ar+ar^(2)+ . . .+ar^(n-1)=(a(r^(n)-1))/(r-1),rgt1]`
`rArrS_(n)=(10)/(9)(10^(n)-1)-n`
Put `n=10`
`rArrS_(10)=(10)/(9)(10^(10)-1)-10`
`=(10)/(9)(10^(10)-1-9)`
`=(10)/(9)(10^(10)-10)=(100)/(9)(10^(9)-1)`