`sum_(i=1)^(oo)sum_(j=1)^(oo)sum_(k=1)^(oo)(1)/(a^(i+j+k))` is equal to (where `|a| gt 1`)

Options

A. `(a-1)^(-3)`
B. `(3)/(a-1)`
C. `(3)/(a^(3)-1)`
D. None of these

Answer

A

Explanation

`(a)` `S=sum_(i=1)^(oo)sum_(j=1)^(oo)sum_(k=1)^(oo)(1)/(a^(i+j+k))`, `|a| gt 1` or `0 lt (1)/(|a|) lt 1`
`=sum_(i=1)^(oo)sum_(j=1)^(oo)sum_(k=1)^(oo)(1)/(a^(i)a^(j)a^(k))`
`=(sum_(i=1)^(oo)(1)/(a^(i)))(sum_(j=1)^(oo)(1)/(a^(j)))(sum_(k=1)^(oo)(1)/(a^(k)))`
`=((1)/(a))/(1-(1)/(a))*((1)/(a))/(1-(1)/(a))*((1)/(a))/(1-(1)/(a))=(1)/((a-1)^(3))`