Answer

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**Hint:**

We are given the general formula of ${r^{th}}$ term. Substituting different values for $r$ and comparing with the given terms we get the value of $a, b$ and $c$. Then we can consider the partial sum of the expression and see the infinite sum by tending $n$ to infinity.

**Useful formula:**

For any $a,b$ we have the algebraic identities.

${(a + b)^2} = {a^2} + 2ab + {b^2}$

${a^2} - {b^2} = (a - b)(a + b)$

**Complete step by step solution:**

It is given that the \[{r^{th}}\] term of the series is $\dfrac{{ar}}{{b{r^4} + c}}$

So substituting $r = 1$, we get the first term.

That is, $\dfrac{{a \times 1}}{{b \times {1^4} + c}} = \dfrac{a}{{b + c}}$

But we have, first term given as $\dfrac{4}{5}$.

This gives, $\dfrac{a}{{b + c}} = \dfrac{4}{5}$.

So we have, $a = 4,b + c = 5$

Now substitute $r = 2$,

$\dfrac{{a \times 2}}{{b \times {2^4} + c}} = \dfrac{{2a}}{{16b + c}}$

But we have, second term as $\dfrac{8}{{65}}$.

Comparing we get,

$16b + c = 65$

Also $b + c = 5$

Subtracting both sides we get,

$15b = 60$

Dividing both sides by $15$ we get,

$b = \dfrac{{60}}{{15}} = 4$

$ \Rightarrow c = 5 - b = 5 - 4 = 1$

So we have, $a = 4,b = 4,c = 1$

Therefore the ${r^{th}}$ term becomes $\dfrac{{4r}}{{4{r^4} + 1}}$.

Consider $4{r^4} + 1$.

Adding and subtracting $4{r^2}$ in the right side we get,

$4{r^4} + 1 = 4{r^4} + 4{r^2} + 1 - 4{r^2}$

Now we have,

$4{r^4} + 1 = (4{r^4} + 4{r^2} + 1) - 4{r^2}$

We have ${(a + b)^2} = {a^2} + 2ab + {b^2}$

Using this we get,

$4{r^4} + 1 = {(2{r^2} + 1)^2} - 4{r^2}$

We know, ${a^2} - {b^2} = (a - b)(a + b)$

$ \Rightarrow 4{r^4} + 1 = [(2{r^2} + 1) - 2r][(2{r^2} + 1) + 2r]$

$ \Rightarrow 4{r^4} + 1 = (2{r^2} + 1 - 2r)(2{r^2} + 1 + 2r)$

Then $\dfrac{{4r}}{{4{r^4} + 1}} = \dfrac{{4r}}{{(2{r^2} + 1 - 2r)(2{r^2} + 1 + 2r)}}$

$\dfrac{{4r}}{{4{r^4} + 1}} = \dfrac{1}{{(2{r^2} + 1 - 2r)}} - \dfrac{1}{{(2{r^2} + 1 + 2r)}}$

Now consider the partial sum,

$\sum\limits_1^n {\dfrac{{4r}}{{4{r^4} + 1}}} = \sum\limits_1^n {(\dfrac{1}{{(2{r^2} + 1 - 2r)}} - \dfrac{1}{{(2{r^2} + 1 + 2r)}}} )$

Substituting values $r = 1,2,3...$ we get,

\[\sum\limits_1^n {\dfrac{{4r}}{{4{r^4} + 1}}} = \dfrac{1}{{2 + 1 - 2}} - \dfrac{1}{{2 + 1 + 2}} + \dfrac{1}{{2 + 1 + 2}} - \dfrac{1}{{8 + 1 + 4}} + \dfrac{1}{{8 + 1 + 4}} - ...\]

\[\sum\limits_1^n {\dfrac{{4r}}{{4{r^4} + 1}}} = 1 - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{{13}} + \dfrac{1}{{13}} - ...\]

So we can see that the consecutive terms get cancelling.

When $n \to \infty $, \[\sum\limits_1^n {\dfrac{{4r}}{{4{r^4} + 1}}} = 1 - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{{13}} + \dfrac{1}{{13}} - ... \to 1\]

So we get the sum up to infinity of the series is $1$.

**$\therefore $ The answer is option D.**

**Note:**

This problem includes a lot of steps. So we may have made mistakes easily. Also remember that when $n \to \infty $, then $\dfrac{1}{n} \to 0$ and so $1 - \dfrac{1}{n} \to 1$. We often use the idea of partial sum in finding the infinite sum.

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