Question

Let $A_n=\left(\frac{3}{4}\right)-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3{\left(-1\right)}^{n-1}{\left(\frac{3}{4}\right)}^n$ and $B_n=1-A_n.$ Then, the least odd natural number p, so that $B_n>A_n$, for all $n\ge p,$ is :

• A. 9
• B. 7
• C. 11
• D. 5

Answer 1

Answer: (B)

Given: $A_n=\left(\frac{3}{4}\right)-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3-\cdots +(-1{)}^{n-1}{\left(\frac{3}{4}\right)}^n$
$A_n=(3/4)\frac{\left(1-(-3/4{)}^n\right)}{(1+3/4)}=\frac{3}{7}\left(1-{\left(\frac{-3}{4}\right)}^n\right)$
$B_n>A_n\Rightarrow 1-A_n>A_n\Rightarrow A_n<\frac{1}{2}\Rightarrow \frac{3}{7}\left(1-{\left(\frac{-3}{4}\right)}^n\right)<\frac{1}{2}$
$\Rightarrow 1-{\left(\frac{-3}{4}\right)}^n<\frac{7}{6}\Rightarrow {\left(\frac{-3}{4}\right)}^n>\frac{-1}{6}\Rightarrow n>6.228\Rightarrow n=7$