Question

Let $A_n=\frac{3}{4}-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3-....+(-1{)}^{n-1}{\left(\frac{3}{4}\right)}^n$
and $B_n=1-A_n,$ then find the least value of $n_0,n_0\in N$ such that $B_n>A_n,\forall n\ge n_0$.

Answer 1

Given:

$A_n=\frac{3}{4}-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3-....+(-1{)}^{n-1}{\left(\frac{3}{4}\right)}^n$

Clearly this series in G.P.

Here, $a=\frac{3}{4}$

$r=\frac{(\frac{3}{4}{)}^2}{\left(\frac{3}{4}\right)}=\frac{3}{4}<1$

$A_n=$ sum of G.P. of $n$ terms whose common radio is $-\frac{3}{4}$

$A_n=\frac{a(1-r^n)}{1-r}$
$\therefore A_n=\frac{3}{4}\frac{[1-{(-\frac{3}{4})}^n]}{1-(-\frac{3}{4})}$

$=\frac{3}{4}.\frac{4}{7}.[1-{(-\frac{3}{4})}^n]$

$=\frac{3}{7}[1-{(-\frac{3}{4})}^n]$
$B_n=1-A_n$ and

$B_n>A_n\Rightarrow 1-A_n>A_n$
$\Rightarrow 1>2A_n$

$\Rightarrow A_n<\frac{1}{2}$
$\therefore \frac{3}{7}[1-{(-\frac{3}{4})}^n]<\frac{1}{2}$

$\Rightarrow 1-{(-\frac{3}{4})}^n<\frac{7}{6}$
$\Rightarrow 1-\frac{7}{6}<{(-\frac{3}{4})}^n$

$\Rightarrow -\frac{1}{6}<{(-\frac{3}{4})}^n$
$\Rightarrow {(-\frac{3}{4})}^n>-\frac{1}{6}$
Which is possible only for all $n\ge 6$.