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**Hint:**Here, the sum of $n$ consecutive terms of an expression is given as $\sum\limits_{r = 1}^n {{t_r}} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8}$. We have to firstly find the term ${t_r}$ by using formula ${t_r} = {S_n} - {S_{n - 1}}$. write $\dfrac{1}{{{t_r}}}$ and arrange them in suitable form then apply $\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} $ and we will get that the successive terms cancel each others.

**Complete step-by-step solution:**

Here, it is given that ${S_n} = \sum\limits_{r = 1}^n {{t_r}} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8}$ .

Now, we can find the term ${t_n}$ using the above given formula.

$

\Rightarrow {t_n} = {S_n} - {S_{n - 1}} \\

\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8} - \dfrac{{\left( {n - 1} \right)\left( {n + 1 - 1} \right)\left( {n + 2 - 1} \right)\left( {n + 3 - 1} \right)}}{8} \\

\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{8} - \dfrac{{\left( {n - 1} \right)n\left( {n + 1} \right)\left( {n + 2} \right)}}{8}

$

Here, it is clearly visible that $\dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{8}$ is present in both the terms so, we can take this as common and we can write,

$

\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{8}\left( {\left( {n + 3} \right) - \left( {n - 1} \right)} \right) \\

\Rightarrow {t_n} = \dfrac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{2}

$

So, ${r^{th}}$ term of the required expression is ${t_r} = \dfrac{{r\left( {r + 1} \right)\left( {r + 2} \right)}}{2}$.

Now, we have to find the value of $\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} $. So, firstly write,

$ \Rightarrow \dfrac{1}{{{t_r}}} = \dfrac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}$

We have to add and subtract $2$ in the numerator of the term $\dfrac{1}{{{t_r}}}$ so that this can be converted into suitable formats.

$ \Rightarrow \dfrac{1}{{{t_r}}} = \dfrac{{\left( {r + 2} \right) - r}}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}$

We can write this as the difference of two fractions. That is

$ \Rightarrow \dfrac{1}{{{t_r}}} = \dfrac{1}{{r\left( {r + 1} \right)}} - \dfrac{1}{{\left( {r + 1} \right)\left( {r + 2} \right)}}$

So, $\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} = \dfrac{1}{{r\left( {r + 1} \right)}} - \dfrac{1}{{\left( {r + 1} \right)\left( {r + 2} \right)}}$

Here, we have to find the summation of $n$ terms, so we have to put the value of $r$ from $1$ to $n$.

$\sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} = \dfrac{1}{{1\left( {1 + 1} \right)}} - \dfrac{1}{{\left( {1 + 1} \right)\left( {1 + 2} \right)}} + \dfrac{1}{{2 \times 3}} - \dfrac{1}{{3 \times 4}} - - - - - - - - + \dfrac{1}{{n\left( {n + 1} \right)}} - \dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$

It is clearly visible that ${2^{nd}}$ and ${3^{rd}}$ terms are cancelling each other and similarly next two terms cancel each other and finally only the first and last terms will be remaining. This imply

$ \Rightarrow \sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}} = \dfrac{1}{2} - \dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}} $

Taking $ - 1$ as common we can write,

$ \Rightarrow \sum\limits_{r = 1}^n {\dfrac{1}{{{t_r}}}} = - \left( {\dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}} - \dfrac{1}{2}} \right)$

**Thus, option (A) is the correct answer.**

**Note:**While solving the problem of summation of sequences and series we have to first write ${r^{th}}$ term and then convert this into suitable form so that except some terms others are cancelled out.

If the denominator of ${r^{th}}$term is cubic like \[\dfrac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}}\] we can write

$ \Rightarrow \dfrac{2}{{r\left( {r + 1} \right)\left( {r + 2} \right)}} = \dfrac{A}{n} + \dfrac{B}{{n + 1}} + \dfrac{C}{{n + 2}}$ and by equating on both side of equation we can get the value of $A,B$ and $C$ and then do summation as shown above.

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