about to only mathematics

Options

A. 5
B. 6
C. 7
D. none of these

Answer

B

Explanation

We have
`A_(n)=(3)/(4)-((3)/(4))^(2)+((3)/(4))^(3)+ . . . .+(-1)^(n-1)((3)/(4))^(n)`
`rArr" "A_(n)=-(-(3)/(4))-((-3)/(4))-((-3)/(4))^(3)+ . . . .+(-)^(n)((-3)/(4))^(n)`
`rArr" "A_(n)=-{:[((-3)/(4)){(1-(-(3)/(4))^(4))/(1-(-(3)/(4)))}]:}=(3)/(7){1-(-(3)/(4))^(n)}`
Now,
`rArr" "B_(n)gtA_(n)`
`rArr" "1-A_(n)gtA_(n)" "[becauseB_(n)=1-A_(n)]`
`rArr" "A_(n)lt(1)/(2)`
`rArr" "(3)/(7){1-(-(3)/(4))^(n)}lt(1)/(2)`
`rArr" "1-(-(3)/(4))^(n)lt(7)/(6)rArr(-(3)/(4))^(n)gt-(1)/(6)`
Clearly, it is true for all even natural numbers.
If n is odd natural number, then
`(-(3)/(4))^(n)gt-(1)/(6)rArr-6xx3^(n)gt-4^(n)rArr3^(n+1)lt2^(2n-1)`
This is true for all `nlt7`. Hence, `n_(0)=7`.