Evaluate the following
$\dfrac{1}{1.5}+\dfrac{1}{5.9}+..........+\dfrac{1}{\left( 4n-3 \right)\left( 4n+1 \right)}=?$

Hint: This is a summation question. We should simplify this and then add. Simplifying here means finding a pattern. For this kind of question , where it goes until infinitely , we are supposed to find it’s ${{n}^{th}}$ term and then add. But here , we are provided with the ${{n}^{th}}$ in the question. So we are just supposed to add it. We add them using the method of differences.

Complete step by step answer:
Now, we can write \[\dfrac{1}{1.5}+\dfrac{1}{5.9}+..........+\dfrac{1}{\left( 4n-3 \right)\left( 4n+1 \right)}=\sum\limits_{r=1}^{n}{\dfrac{1}{\left( 4r-3 \right)\left( 4r+1 \right)}}\]
Where r is a random variable and is not connected to the question whatsoever. We only considered so as to simplify the question and get the answer easily.
We will solve this using the method of differences.
We are only trying to shrink our question.
$\Rightarrow \sum\limits_{r=1}^{n}{\dfrac{1}{\left( 4r-3 \right)\left( 4r+1 \right)}}$ when we expand this , we get our question back.
So now , we should use the concept of partial fractions. We have to split our denominator.
We can do it in the following way :
$\sum\limits_{r=1}^{n}{\dfrac{1}{\left( 4r-3 \right)\left( 4r+1 \right)}}$ is same as $\dfrac{1}{4}\sum\limits_{r=1}^{n}{\dfrac{\left( 4r+1 \right)-\left( 4r-3 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)}}$ .
$\Rightarrow \sum\limits_{r=1}^{n}{\dfrac{1}{\left( 4r-3 \right)\left( 4r+1 \right)}}=\dfrac{1}{4}\sum\limits_{r=1}^{n}{\dfrac{\left( 4r+1 \right)-\left( 4r-3 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)}}$
Now, we have to just solve this $\dfrac{1}{4}\sum\limits_{r=1}^{n}{\dfrac{\left( 4r+1 \right)-\left( 4r-3 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)}}$.
\[\Rightarrow \dfrac{1}{4}\sum\limits_{r=1}^{n}{\dfrac{\left( 4r+1 \right)-\left( 4r-3 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)}}=\dfrac{1}{4}\sum\limits_{r=1}^{n}{{}}\dfrac{\left( 4r+1 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)}-\dfrac{\left( 4r-3 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)}\]
So, now we just have to solve this \[\dfrac{1}{4}\sum\limits_{r=1}^{n}{{}}\dfrac{\left( 4r+1 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)}-\dfrac{\left( 4r-3 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)}\].
\[\begin{align}
  & \Rightarrow \dfrac{1}{4}\sum\limits_{r=1}^{n}{{}}\dfrac{\left( 4r+1 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)}-\dfrac{\left( 4r-3 \right)}{\left( 4r-3 \right)\left( 4r+1 \right)} \\
 & \Rightarrow \dfrac{1}{4}\sum\limits_{r=1}^{n}{\dfrac{1}{\left( 4r-3 \right)}-\dfrac{1}{\left( 4r+1 \right)}} \\
\end{align}\]
This is called the method of differences.
Now let’s expand back . Upon expanding , we get the following :
$\Rightarrow \dfrac{1}{4}\left[ 1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}.........+\dfrac{1}{4n-4}-\dfrac{1}{4n-3}+\dfrac{1}{4n-3}-\dfrac{1}{4n+1} \right]$
Everything will get cancelled except $1\And \dfrac{1}{4n+1}$ .
 We get :
$\Rightarrow \dfrac{1}{4}\left[ 1-\dfrac{1}{4n+1} \right]$
Let us take L.C.M. We get the following :
\[\begin{align}
  & \Rightarrow \dfrac{1}{4}\left[ 1-\dfrac{1}{4n+1} \right] \\
 & \Rightarrow \dfrac{1}{4}\left[ \dfrac{4n+1-1}{4n+1} \right] \\
 & \Rightarrow \dfrac{1}{4}\left[ \dfrac{4n}{4n+1} \right] \\
 & \Rightarrow \dfrac{n}{4n+1} \\
\end{align}\]
So the answer is \[\dfrac{n}{4n+1}\] .
$\therefore $ Hence , upon solving $\dfrac{1}{1.5}+\dfrac{1}{5.9}+..........+\dfrac{1}{\left( 4n-3 \right)\left( 4n+1 \right)}$ , we get \[\dfrac{n}{4n+1}\].

Additional information:
We can even verify our answers. Just plug – in the value of n = 1 , 2 , 3 ….. and see whether we are getting the value of the sum of the first two terms in the series when n = 2 is plugged or the value of the sum of the first three terms when n = 3 is plugged.

Note:
In these kinds of questions , a pattern must be observed. And this is only achieved when there is practice of different kinds of summation questions. The method of difference is very important for solving these kinds of questions. We have to be very careful while solving these kinds of questions as there is a lot of scope of calculation errors.