If `a_n=3/4-(3/4)^2+(3/4)^3+...(-1)^(n-1)(3/4)^n` and `b_n=1-a_n`, then find the minimum natural number n, such that `b_n>a_n`

Explanation

`:.B_(n)=1-A_(n)gtA_(n)implies A_(n)lt(1)/(2)`
Now, `A_(n)=((3)/(4)(1-(-(3)/(4))^(n)))/(1+(3)/(4))lt(1)/(2) implies (-(3)/(4))^(n)gt -(1)/(6)`
Obviously, it is true for all even values of n.
But for `n=1,-(3)/(4)lt - (1)/(6)`
`n=3(-(3)/(4))^(3)=-(27)/(24)lt - (1)/(6)`
`n=5,(-(3)/(4))^(5)= -(243)/(1024)gt -(1)/(6)`
which is true for n=7 obviously, `n_(0)=7`
Aliter `B_(n)=1-A_(n)gtA_(n)`
`implies A_(n)lt(1)/(2)implies (3)/(4)((1-(-(3)/(4))^(n)))/(1+(3)/(4))lt(1)/(2) implies (-(3)/(4))^(n)gt -(1)/(6)`
Obviously, it is true for all even values of n.
But for `n=1,-(3)/(4)lt -(1)/(6)`
`n=3,(-(3)/(4))^(3)=-(27)/(64)lt - (1)/(6)`
`n=5,(-(3)/(4))^(5)= -(243)/(1024)lt -(1)/(6)`
and for `n=7 implies (-(3)/(4))^(7)=-(2187)/(12288)gt -(1)/(6)`
Hence, minimum natural number `n_(0)=7`.