Question

The sum of the series $\frac{9}{5^2.2.1}+\frac{13}{5^3.3.2}+\frac{17}{5^4.4.3}+...$ to infinite terms, is

• A. $\frac{2}{5}$
• B. $\frac{1}{5}$
• C. $1$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{5}$

The general term of the series is

$t_r=\frac{4r+1}{5^r.r(r-1)}$, where $r\ge z$

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

$\therefore \sum _{r=2}^{\infty }{t}_r=(\frac{1}{5^1.1}-\frac{1}{5^2.2})+(\frac{1}{5^{2}2}-\frac{1}{5^{3}3})+(\frac{1}{5^{3}3}-\frac{1}{5^{4}4})+...$ to infinity

$=\frac{1}{5}(\because$ terms tend to zero an $n\to \infty )$