Question

For a sequence if $S_n=\frac{4^n-3^n}{3^n}$ find the nth term hence show that if it is a G.P

Answer 1

Given,

$S_n=\frac{4^n-3^n}{3^n}$

$\Rightarrow S_n={\left(\frac{4}{3}\right)}^n-1$

$S_{n-1}={\left(\frac{4}{3}\right)}^{n-1}-1$

$T_n=S_n-S_{n-1}$

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

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

$\therefore T_n=\frac{1}{4}{\left(\frac{4}{3}\right)}^n$

$T_1=\frac{1}{4}{\left(\frac{4}{3}\right)}^1=\frac{1}{3}$

$T_2=\frac{1}{4}{\left(\frac{4}{3}\right)}^2=\frac{4}{9}$

$T_3=\frac{1}{4}{\left(\frac{4}{3}\right)}^3=\frac{16}{27}$

$\frac{T_2}{T_1}=\frac{4}{9}\times \frac{3}{1}=\frac{4}{3}$

$\frac{T_3}{T_2}=\frac{16}{27}\times \frac{9}{4}=\frac{4}{3}$

As $\frac{T_2}{T_1}=\frac{T_3}{T_2}$, given series is in G.P