Question

The sum of the series
$\frac{9}{5^2\cdot 2.1}+\frac{13}{5^3\cdot 3.2}+\frac{17}{5^4\cdot 4.3}..........$ Infinite terms

• A. $\frac{1}{25}$
• B. $\frac{1}{5}$
• C. Not a finite number
• D. $\frac{9}{5}$

Answer 1

The correct option is B $\frac{1}{5}$

$\frac{9}{5^2\cdot 2\cdot 1}+\frac{13}{5^3\cdot 3\cdot 2}+\frac{17}{5^4\cdot 4\cdot 3}................\infty \text{term}$

The general term of the above series,

for, $r⩾2$

General term of the numerator

$T_r^{\prime }=4r+1$

General term of the denominater

$T_{r^{\prime \prime }}=\frac{1}{r(r-1)}$

Then

$T_r=\frac{4r+1}{r(r-1)\cdot 5^r}$ where, $r⩾2$

$=\frac{5r-(r-1)}{5^{r}r(r-1)}$

$=\frac{5r}{5^r\cdot \left(r\right(r-1)}-\frac{r-1}{r\cdot (r-1)}$

$\Rightarrow \frac{1}{5^{r-1}(r-1)}-\frac{1}{5^r\cdot r}$

Now, putting the $r⩾2$ in above equation

$T_2=\frac{1}{5\cdot 1}-\frac{1}{5^2\cdot 2}$

$T_3=\frac{1}{5^2\cdot 2}-\frac{1}{5^3\cdot 3}$

$\because \text{every last term is gettimg cancelled}$

$\therefore \text{o}{\sum }_{r=2}^{\infty }{t}_r=\frac{1}{5}$

Option B