Question

Suppose a series of $n$ terms is given by $S_n=a_1+a_2+...+a_n$ Then $S_{n-1}=a_1+a_2+...+a_{n-1}(\forall n>1)$ and then we can write $t_n=S_n-S_{n-1}\forall n\ge 2$ for $n=1$ we have $S_1=t_1$
On the basis of above information answer the following questionsIf the sum to n terms of a series is $a{3}^n-b$ then $\sum _{r=2}^{\infty }\frac{1}{t_r}$ is equal to

• A. $\frac{1}{4a}$
• B. $\frac{1}{2a}$
• C. $\frac{2}{3a}$
• D. $\frac{3}{2a}$

Answer 1

The correct option is A $\frac{1}{4a}$

Given that, $S_n=a{3}^n-b$
$\Rightarrow S_{n-1}=a{3}^{n-1}-b$
$\therefore t_n=2a{3}^{n-1}$
$\therefore \sum _{r=2}^{\infty }\frac{1}{t_r}=\frac{1}{2a}\sum _{r=2}^{\infty }\left(\frac{1}{3^{r-1}}\right)=\frac{1}{2a}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...\infty )=\frac{1}{2a}\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)$
$\Rightarrow {\sum }_{r=2}^{\infty }\frac{1}{t_r}=\frac{1}{4a}$