Question

If $\sum _{r=1}^{n}l\left(r\right)=3^n-1$, then $\sum _{r=1}^n\frac{1}{l\left(r\right)}$ is equal to?

• A. $2(1-{\left(\frac{1}{3}\right)}^n)$
• B. $3(1-{\left(\frac{1}{3}\right)}^n)$
• 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 A $2(1-{\left(\frac{1}{3}\right)}^n)$

Given,

${\sum }_{r=1}^{n}l\left(r\right)=T_1+T_2+......+T_n$

${\sum }_{r=1}^{n-1}l\left(r\right)=T_r=T_1+T_2+......+T_{n-1}$

${\sum }_{r=1}^n{T}_r-{\sum }_{r=1}^{n-1}{T}_r=T_n$

$⟹T_r=(3^n-1)-(3^{n-1}-1)$

$⟹T_r=3^n-\frac{3^n}{3}-1+1$

$⟹T_r=\frac{3}{2}\times \frac{1}{3^r}$

$\therefore {\sum }_{r=1}^n=\frac{3}{2}(\frac{1}{3}+\frac{1}{3^2}+......)$

$=\frac{3}{2}.\frac{\frac{1}{3}(1-(\frac{1}{3}{)}^n}{1-\frac{1}{3}}$

$=2(1-(\frac{1}{3}{)}^n)$