First slide

Special Series in Sequences and Series

Question

$If \sum_{r = 1}^{n} T_{r} = (3^{n} - 1), then find the sum of \sum_{m = 1}^{n} \frac{1}{T_{m}} .$

Moderate

A

$\frac{3}{4} (\frac{1}{3^{n}})$
B

$\frac{3}{4} (1 - \frac{1}{3^{n}})$
C

$\frac{3}{4} (1 - \frac{1}{2^{n}})$
D

$\frac{3}{4} (1 - \frac{1}{5^{n}})$

Solution

$\sum_{r = 1}^{n} T_{r} = 3^{n} - 1$
$\begin{aligned} \Rightarrow T_{r} = (3^{r} - 1) - (3^{r - 1} - 1) = 3^{r - 1} (3 - 1) = 2 (3^{r - 1}) \\ \Rightarrow \frac{1}{T_{r}} = \frac{1}{2} \times {(\frac{1}{3})}^{r - 1} \end{aligned}$
$\begin{array}{l} \Rightarrow \sum_{r = 1}^{n} \frac{1}{T_{r}} = \frac{1}{2} \sum_{r = 1}^{n} {(\frac{1}{3})}^{r - 1} \\ = \frac{1}{2} \frac{{(1 - \frac{1}{3})}^{n}}{1 - \frac{1}{3}} \\ = \frac{3}{4} (1 - {(\frac{1}{3})}^{n}) \end{array}$

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