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 $n \geq p,$ is :

• A. 9
• B. 7
• C. 11
• D. 5

Answer 1

Answer: (B)

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