Question

If $\sum _{r=1}^n{T}_r=(3^n-1),$ then the value of $\sum _{r=1}^n\frac{1}{T_r}$ is

• A. $\frac{4}{3}(1-3^n)$
• B. $\frac{3}{2}[1-{\left(\frac{1}{3}\right)}^n]$
• C. $\frac{3}{4}[1-{\left(\frac{1}{3}\right)}^n]$
• D. $\frac{4}{3}[1-{\left(\frac{1}{3}\right)}^n]$

Answer 1

The correct option is C $\frac{3}{4}[1-{\left(\frac{1}{3}\right)}^n]$

$\sum _{r=1}^n{T}_r=3^n-1$
$\Rightarrow T_r=(3^r-1)-(3^{r-1}-1)=3^{r-1}(3-1)=2\left(3^{r-1}\right)$

$\Rightarrow \frac{1}{T}=\frac{1}{2}\times {\left(\frac{1}{3}\right)}^{r-1}$
$\Rightarrow \sum _{r=1}^n\frac{1}{T_r}=\frac{1}{2}\sum _{r=1}^n{\left(\frac{1}{3}\right)}^{r-1}$
$=\frac{1}{2}\frac{(1-{\left(\frac{1}{3}\right)}^n)}{1-\frac{1}{3}}$
$\therefore \sum _{r=1}^n\frac{1}{T_r}=\frac{3}{4}(1-{\left(\frac{1}{3}\right)}^n)$


Alternate method :
Putting $n=1$,
$T_1=3^1-1=2\Rightarrow \frac{1}{T_1}=\frac{1}{2}$
Now, putting $n=1$ in the given options,
$\frac{4}{3}(1-3^1)=-\frac{8}{3}$
$\frac{3}{2}[1-{\left(\frac{1}{3}\right)}^1]=1$
$\frac{3}{4}[1-{\left(\frac{1}{3}\right)}^1]=\frac{1}{2}$
$\frac{4}{3}[1-{\left(\frac{1}{3}\right)}^1]=\frac{8}{9}$

So, the correct choice is $\frac{3}{4}[1-{\left(\frac{1}{3}\right)}^n]$