The value of ∑r=1∞(4r+5)5−rr(5r+5) is

∑r=1∞((4r+5)5−rr(5r+5)).15r=∑r=1∞(1r−15r+5).15r=∑r=1∞(1r.5r−1(r+1).5r+1)=15.

The value of ∑r=1∞(4r+5)5−rr(5r+5) is

∑r=1∞((4r+5)5−rr(5r+5)).15r=∑r=1∞(1r−15r+5).15r=∑r=1∞(1r.5r−1(r+1).5r+1)=15.

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The value of $\sum_{r = 1}^{\infty} \frac{(4 r + 5) 5^{- r}}{r (5 r + 5)}$ is

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$\frac{2}{25}$

$\frac{2}{5}$

$\frac{1}{25}$

$\frac{1}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} \sum_{r = 1}^{\infty} (\frac{(4 r + 5) 5^{- r}}{r (5 r + 5)}) . \frac{1}{5^{r}} \\ = \sum_{r = 1}^{\infty} (\frac{1}{r} - \frac{1}{5 r + 5}) . \frac{1}{5^{r}} \\ = \sum_{r = 1}^{\infty} (\frac{1}{r {.5}^{r}} - \frac{1}{(r + 1) {.5}^{r + 1}}) = \frac{1}{5} . \end{array}$