In the given series $\dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+...+\text{up to}~n\text{ terms}.$
Find the sum of n terms.

Hint: To find the sum of n terms, first of all we will find the ${{n}^{th}}$ term of the given equation. When we will get the ${{n}^{th}}$ term of the given equation, we will use a summation sign in every term of the ${{n}^{th}}$ term. Since, we used a summation sign that means we got the sum of the given equation up to n terms but if there is any possibility to simplify it, we will get the simplified sum of the given question up to n terms.

Complete step-by-step solution:
Since, the given question is:
$\Rightarrow \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+...+\text{up to}~n\text{ terms}.$
First of all we will get the ${{n}^{th}}term$ of the given equation as:
$\Rightarrow {{n}^{th}}term=\dfrac{1}{\left( 3n-1 \right)\left( 3n+2 \right)\left( 3n+5 \right)}$
Now, we will elaborate it by expanding the above equation. When we expand $\left( 3n-1 \right)$ and $\left( 3n+5 \right)$ so that we will get $6$ as a result. So, there we will divide by $3$ in the above equation as:
$\Rightarrow {{n}^{th}}term=\dfrac{1}{\left( 3n+2 \right)}\dfrac{1}{6}\left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right)$
After that we will open the bracket and will multiply $\left( 3n-1 \right)$to the terms that are inside the bracket as:
$\Rightarrow {{n}^{th}}term=\dfrac{1}{6}\left( \dfrac{1}{\left( 3n-1 \right)\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+2 \right)\left( 3n+5 \right)} \right)$
Now, again we will separate the in bracketed terms that are in multiplication form as:
$\Rightarrow {{n}^{th}}term=\dfrac{1}{3}\left[ \left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)} \right)-\left( \dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right) \right]$
Since, we will get $3$when we will use subtraction in both the bracketed terms. So we will divide them by $3$ as:
$\Rightarrow {{n}^{th}}term=\dfrac{1}{6}\left[ \dfrac{1}{3}\left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)} \right)-\dfrac{1}{3}\left( \dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right) \right]$
Now, we will take out $\dfrac{1}{3}$ as a common factor and will multiply it by $\dfrac{1}{6}$ as:
$\Rightarrow {{n}^{th}}term=\dfrac{1}{6}\times \dfrac{1}{3}\left[ \left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)} \right)-\left( \dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right) \right]$
$\Rightarrow {{n}^{th}}term=\dfrac{1}{18}\left[ \left( \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)} \right)-\left( \dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+5 \right)} \right) \right]$
Here, we will solve the bracketed terms by opening the bracket as:
$\Rightarrow {{n}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3n-1 \right)}-\dfrac{1}{\left( 3n+2 \right)}-\dfrac{1}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]$
Now, the above equation will be as:
$\Rightarrow {{n}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3n-1 \right)}-\dfrac{2}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]$
Now, here we will try to find out the value of ever term as:
For first term: $n=1$
$\Rightarrow {{1}^{st}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3\times 1-1 \right)}-\dfrac{2}{\left( 3\times 1+2 \right)}+\dfrac{1}{\left( 3\times 1+5 \right)} \right]$
$\Rightarrow {{1}^{st}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3-1 \right)}-\dfrac{2}{\left( 3+2 \right)}+\dfrac{1}{\left( 3+5 \right)} \right]$
$\Rightarrow {{1}^{st}}term=\dfrac{1}{18}\left[ \dfrac{1}{2}-\dfrac{2}{5}+\dfrac{1}{8} \right]$
For second term: $n=2$
$\Rightarrow {{2}^{nd}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3\times 2-1 \right)}-\dfrac{2}{\left( 3\times 2+2 \right)}+\dfrac{1}{\left( 3\times 2+5 \right)} \right]$
After solving above equation we will get:
$\Rightarrow {{2}^{nd}}term=\dfrac{1}{18}\left[ \dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{11} \right]$
For third term: $n=3$
$\Rightarrow {{3}^{rd}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3\times 3-1 \right)}-\dfrac{2}{\left( 3\times 3+2 \right)}+\dfrac{1}{\left( 3\times 3+5 \right)} \right]$
We will have the value of third term after solving the above equation as:
$\Rightarrow {{3}^{rd}}term=\dfrac{1}{18}\left[ \dfrac{1}{8}-\dfrac{2}{11}+\dfrac{1}{14} \right]$
… … … … …
For ${{\left( n-1 \right)}^{th}}term$ : $n=n-1$
$\Rightarrow {{\left( n-1 \right)}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{3\left( n-1 \right)-1}-\dfrac{2}{3\left( n-1 \right)+2}+\dfrac{1}{3\left( n-1 \right)+5} \right]$
Now, the ${{\left( n-1 \right)}^{th}}term$ will be as:
$\Rightarrow {{\left( n-1 \right)}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{3n-4}-\dfrac{2}{3n-1}+\dfrac{1}{3n+2} \right]$
This process will go up to ${{n}^{th}}term$.
Since, we got all the $n\text{ }terms$ of the given question. Now, we will apply summation sign to get the sum of the given question as:
\[\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \sum\limits_{1}^{n}{\dfrac{1}{\left( 3n-1 \right)}}-\sum\limits_{1}^{n}{\dfrac{2}{\left( 3n+2 \right)}}+\sum\limits_{1}^{n}{\dfrac{1}{\left( 3n+5 \right)}} \right]\]
Now, we will write it as:
$\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{1}{2}-\dfrac{2}{5}+\dfrac{1}{8} \right]+\dfrac{1}{18}\left[ \dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{11} \right]+\dfrac{1}{18}\left[ \dfrac{1}{8}-\dfrac{2}{11}+\dfrac{1}{14} \right]+... \\
 & +\dfrac{1}{18}\left[ \dfrac{1}{3n-4}-\dfrac{2}{3n-1}+\dfrac{1}{3n+2} \right]+\dfrac{1}{18}\left[ \dfrac{1}{\left( 3n-1 \right)}-\dfrac{2}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right] \\
\end{align}$
$\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{1}{2}-\dfrac{2}{5}+\dfrac{1}{8}+\dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{11}+\dfrac{1}{8}-\dfrac{2}{11}+\dfrac{1}{14}+...+\dfrac{1}{3n-4}-\dfrac{2}{3n-1}+\dfrac{1}{3n+2}+\dfrac{1}{\left( 3n-1 \right)}-\dfrac{2}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]$
As we can see that some terms will be canceled. So, we will get the above equation as:
$\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{1}{2}-\dfrac{1}{5}-\dfrac{1}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]$
Now, we will solve the above equation as:
$\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{5-2}{10}-\left( \dfrac{\left( 3n+5 \right)-\left( 3n+2 \right)}{\left( 3n+2 \right)\left( 3n+5 \right)} \right) \right]$
$\Rightarrow {{S}_{n}}=\dfrac{1}{18}\left[ \dfrac{3}{10}-\dfrac{3}{\left( 3n+2 \right)\left( 3n+5 \right)} \right]$
Here, we will take $3$ as common:
$\Rightarrow {{S}_{n}}=\dfrac{3}{18}\left[ \dfrac{1}{10}-\dfrac{1}{\left( 3n+2 \right)\left( 3n+5 \right)} \right]$
Since, we got a sum of the n terms of the given question but we can simplify it further. So, we will use subtraction to solve it as:
$\Rightarrow {{S}_{n}}=\dfrac{1}{6}\left[ \dfrac{\left( 3n+2 \right)\left( 3n+5 \right)-10}{10\left( 3n+2 \right)\left( 3n+5 \right)} \right]$
Now, we will expand the above equation by opening bracket as:
$\Rightarrow {{S}_{n}}=\dfrac{1}{6}\left[ \dfrac{9{{n}^{2}}+6n+15n+10-10}{10\left( 9{{n}^{2}}+6n+15n+10 \right)} \right]$
Now, we will do the necessary calculation as:
$\Rightarrow {{S}_{n}}=\dfrac{1}{6}\left[ \dfrac{9{{n}^{2}}+21n}{\left( 90{{n}^{2}}+210n+100 \right)} \right]$
$\Rightarrow {{S}_{n}}=\dfrac{1}{6}\left[ \dfrac{3\left( 3{{n}^{2}}+7n \right)}{\left( 90{{n}^{2}}+210n+100 \right)} \right]$
$\Rightarrow {{S}_{n}}=\dfrac{3}{6}\left[ \dfrac{n\left( 3n+7 \right)}{\left( 90{{n}^{2}}+210n+100 \right)} \right]$
$\Rightarrow {{S}_{n}}=\dfrac{1}{2}\left[ \dfrac{n\left( 3n+7 \right)}{\left( 90{{n}^{2}}+210n+100 \right)} \right]$
Hence, this is a simplified solution of the sum of n terms of the given equation.

Note: Here, we will use alternative method to check the sum of the given question as:
Since, we will have to find the sum of the given question for infinite terms that is:
$\Rightarrow \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+...$
Here, we will consider the sum of some terms as:
$\Rightarrow \dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...\text{ = }{{S}_{\infty }}$ … $\left( i \right)$
Now, we can change the place of first term and can place them other side of equal sign as:
$\Rightarrow \dfrac{1}{5.8}+\dfrac{1}{8.11}+...\text{ = }{{S}_{\infty }}-\dfrac{1}{2.5}$ … $\left( ii \right)$
Since, we have two equations; we need to subtract equation $\left( i \right)$ from equation $\left( ii \right)$ so that we can solve the question easily as:
\[\begin{align}
  & \underline{\begin{align}
  & \Rightarrow \dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...\text{ = }{{S}_{\infty }} \\
 & \Rightarrow \dfrac{1}{5.8}+\dfrac{1}{8.11}+...\text{ = }{{S}_{\infty }}-\dfrac{1}{2.5} \\
\end{align}} \\
 & \Rightarrow \dfrac{6}{2.5.8}+\dfrac{6}{5.8.11}+\dfrac{6}{8.11.14}...\text{ =}\dfrac{1}{2.5} \\
\end{align}\]
Since, $6$ is a common factor in all the terms of the above equation. So, we can take $6$common as:
\[\Rightarrow 6\left( \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+... \right)\text{ =}\dfrac{1}{2.5}\]
As we have $6$ , a common factor in the above equation. So we can divide by $6$ in the above equation as:
\[\Rightarrow \dfrac{6}{6}\left( \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+... \right)\text{ =}\dfrac{1}{2.5}\times \dfrac{1}{6}\]
Here, $6$ will cancel out $6$ and after that we will have the multiplication of $2$ , $5$ and $6$as
\[\Rightarrow \left( \dfrac{1}{2.5.8}+\dfrac{1}{5.8.11}+\dfrac{1}{8.11.14}+... \right)\text{ =}\dfrac{1}{60}\]
Hence, we have $\dfrac{1}{60}$ as the sum of infinite terms of a given question.
Since, from the solution we have ${{n}^{th}}term$ of the given question as:
$\Rightarrow {{n}^{th}}term=\dfrac{1}{18}\left[ \dfrac{1}{\left( 3n-1 \right)}-\dfrac{2}{\left( 3n+2 \right)}+\dfrac{1}{\left( 3n+5 \right)} \right]$
And the sum of the solution is:
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{18}\left[ \dfrac{3}{10}-\dfrac{3}{\left( 3n+2 \right)\left( 3n+5 \right)} \right]$
Since, we will replace n by infinite. So, the second term of the bracket approximately will be equal to zero. Hence we will have:
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{18}\left[ \dfrac{3}{10}-0 \right]$
Now, we will solve the above equation as:
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{18}\times \dfrac{3}{10}$
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{6}\times \dfrac{1}{10}$
$\Rightarrow {{S}_{\infty }}=\dfrac{1}{60}$
Here, we also get the value of the sum of infinite terms of a given question as $\dfrac{1}{60}$ .
Hence, the solution is correct.