Question

Find the sum of series:
$\dfrac{1}{{1 \cdot 4}} + \dfrac{1}{{4 \cdot 7}} + \dfrac{1}{{7 \cdot 10}} + ... + \dfrac{1}{{\left( {3n - 5} \right)\left( {3n - 2} \right)}}$.

Answer 1

Hint: The series consists of a combination of two different APs in denominator. First identify its general term using the general term of AP $a + \left( {n - 1} \right)d$. Then for finding the value of the series, take the summation of the series and expand it. Then again identify the pattern to simplify it and get the final result.

Complete step-by-step answer:
According to the question, the given series is $\dfrac{1}{{1 \cdot 4}} + \dfrac{1}{{4 \cdot 7}} + \dfrac{1}{{7 \cdot 10}} + ... + \dfrac{1}{{\left( {3n - 5} \right)\left( {3n - 2} \right)}}$.
We can observe that in every term, the denominator consists of multiplication of two numbers of two different APs.
The first AP is 1, 4, 7,….. and the second AP is 4, 7, 10,…..
We know that the general term of an AP is $a + \left( {n - 1} \right)d$, where $a$ is the first term and $d$ is the common difference.
Now, let ${t_n}$ denote the${n^{{\text{th}}}}$ term of the series then using the formula of general term of AP, we have:
$
   \Rightarrow {t_n} = \dfrac{1}{{\left[ {1 + \left( {n - 1} \right) \times 3} \right]\left[ {4 + \left( {n - 1} \right) \times 3} \right]}} \\
   \Rightarrow {t_n} = \dfrac{1}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}} \\
$
Multiplying and dividing by 3 on the RHS, we’ll get:
$ \Rightarrow {t_n} = \dfrac{1}{3}\left[ {\dfrac{3}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}}} \right]$
Now, this expression can further be simplified as:
\[
   \Rightarrow {t_n} = \dfrac{1}{3}\left[ {\dfrac{{\left( {3n + 1} \right) - \left( {3n - 2} \right)}}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}}} \right] \\
   \Rightarrow {t_n} = \dfrac{1}{3}\left[ {\dfrac{1}{{\left( {3n - 2} \right)}} - \dfrac{1}{{\left( {3n + 1} \right)}}} \right] \\
\]
For finding the value of $\dfrac{1}{{1 \cdot 4}} + \dfrac{1}{{4 \cdot 7}} + \dfrac{1}{{7 \cdot 10}} + ... + \dfrac{1}{{\left( {3n - 5} \right)\left( {3n - 2} \right)}}$, we have to find the summation of its general term. So we have the value of the series as:
 \[ \Rightarrow \sum {{t_n}} = \sum\limits_{n = 1}^n {\dfrac{1}{3}\left[ {\dfrac{1}{{\left( {3n - 2} \right)}} - \dfrac{1}{{\left( {3n + 1} \right)}}} \right]} \]
Taking $\dfrac{1}{3}$ outside and putting some values of $n$ to again observe the pattern, we have:
\[ \Rightarrow \sum {{t_n}} = \dfrac{1}{3}\left[ {\left( {\dfrac{1}{1} - \dfrac{1}{4}} \right) + \left( {\dfrac{1}{4} - \dfrac{1}{7}} \right) + \left( {\dfrac{1}{7} - \dfrac{1}{{10}}} \right) + \left( {\dfrac{1}{{10}} - \dfrac{1}{{13}}} \right) + ..... + \left( {\dfrac{1}{{3n - 2}} - \dfrac{1}{{3n + 1}}} \right)} \right]\]
We can see that the second number of each term is getting cancelled by the first number of the next term. So, as a result we will get only the first number of first term and second number of last term to be not cancelled. This is shown below:
\[
   \Rightarrow \sum {{t_n}} = \dfrac{1}{3}\left( {1 - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{10}} + \dfrac{1}{{10}} - \dfrac{1}{{13}} + ..... + \dfrac{1}{{3n - 2}} - \dfrac{1}{{3n + 1}}} \right) \\
   \Rightarrow \sum {{t_n}} = \dfrac{1}{3}\left( {1 - \dfrac{1}{{3n + 1}}} \right) \\
\]
On further simplifying it, we’ll get:
\[
   \Rightarrow \sum {{t_n}} = \dfrac{1}{3}\left( {\dfrac{{3n + 1 - 1}}{{3n + 1}}} \right) \\
   \Rightarrow \sum {{t_n}} = \dfrac{1}{3}\left( {\dfrac{{3n}}{{3n + 1}}} \right) \\
   \Rightarrow \sum {{t_n}} = \dfrac{n}{{3n + 1}} \\
\]

Therefore the value of the series $\dfrac{1}{{1 \cdot 4}} + \dfrac{1}{{4 \cdot 7}} + \dfrac{1}{{7 \cdot 10}} + ... + \dfrac{1}{{\left( {3n - 5} \right)\left( {3n - 2} \right)}}$ is \[\dfrac{n}{{3n + 1}}\]

Note: The general term of the above series as calculated above is ${t_n} = \dfrac{1}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}}$.
This general term can also be simplified by partial fraction as shown below:
$ \Rightarrow \dfrac{1}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}} = \dfrac{A}{{3n - 2}} + \dfrac{B}{{3n + 1}}$
So, by comparing the coefficients on both sides we can determine the value of $A$ and $B$. The final result will be same as above:
 $ \Rightarrow \dfrac{1}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}} = \dfrac{1}{3}\left( {\dfrac{1}{{3n - 2}} - \dfrac{1}{{3n + 1}}} \right)$