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Q.
The sum of the series $\frac{9}{5^{2} \cdot 2 \cdot 1}+\frac{13}{5^{3} \cdot 3 \cdot 2}+\frac{17}{5^{4} \cdot 4 \cdot 3}+\ldots$ to infinite terms, is
Sequences and Series
Solution:
The general term of the series is
$t_{r}=\frac{4 r+1}{5^{r} \cdot r(r-1)}$, where $r \geq z$
$=\frac{5 r-(r-1)}{5^{r} \cdot r(r-1)}=\frac{1}{5^{r-1}(r-1)}-\frac{1}{5^{r} \cdot r}$
$\therefore \displaystyle\sum_{r=2}^{\infty} t_{r}=\left(\frac{1}{5^{1} \cdot 1}-\frac{1}{5^{2} \cdot 2}\right)+\left(\frac{1}{5^{2} \cdot 2}-\frac{1}{5^{3} \cdot 3}\right)+\left(\frac{1}{5^{3} \cdot 3}-\frac{1}{5^{4} \cdot 4}\right)+ … $ to infinity
$=\frac{1}{5} $
$(\because$ terms tend to zero as $n \rightarrow \infty)$