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Find the sum of the following series to n terms: 1.2.4 + 2.3.7 + 3.4.10 + ….

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Hint: In this question use the first, second and third term to find the general format of this series such as in the first number can be simplified as [1 + (1 -1)$ \times $1] . [2 + (2 – 1) $ \times $ 1]. [4 + (3 – 1) $ \times $ 3] use this information to make a better approach toward the solution of the problem

Complete step-by-step answer:
According to the given information we have 1st term 1.2.4 which can be simplified as [1 + (1 -1)$ \times $1]. [2 + (1 – 1) $ \times $ 1]. [4 + (1 – 1) $ \times $ 3] 2nd term 2.3.7 which can be written as [1 + (2 -1)$ \times $1]. [2 + (2 – 1) $ \times $ 1]. [4 + (2 – 1) $ \times $ 3] and the 3rd term 3.4.10 can be written as [1 + (3 -1)$ \times $1]. [2 + (3 – 1) $ \times $ 1]. [4 + (3 – 1) $ \times $ 3]
So by observing the above statement it seems a similar pattern in the given series which is [1 + (n -1)$ \times $1]. [2 + (n – 1) $ \times $ 1]. [4 + (n – 1) $ \times $ 3] here n is the number of term in the series
Let’s simplify the equation [1 + (n -1)$ \times $1]. [2 + (n – 1) $ \times $ 1]. [4 + (n – 1) $ \times $ 3]
[1 + (n -1)$ \times $1]. [2 + (n – 1) $ \times $ 1]. [4 + (n – 1) $ \times $ 3] = [1 + n – 1] [2 + n – 1] [4 + 3n – 3]
$ \Rightarrow $n (n + 1) (3n + 1) = \[3{n^3} + {n^2} + 3{n^2} + n\]
$ \Rightarrow $\[3{n^3} + 4{n^2} + n\]
Therefore Tn = \[3{n^3} + 4{n^2} + n\]
As we know that the sum of any series is given by \[{S_n} = \sum\limits_{n = 1}^n {{T_n}} \]
Substituting the value of Tn in the above formula
\[{S_n} = \sum\limits_{n = 1}^n {{T_n}} = 3\sum\limits_{n = 1}^n {{n^3}} + 4\sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n n \] (Equation 1)
For \[3\sum\limits_{n = 1}^n {{n^3}} \] the terms can be \[{1^3} + {2^3} + {3^3} + \ldots \ldots + {n^3}\]and the sum of these term is given as \[{S_n} = \dfrac{{{{\left( {n\left( {n + 1} \right)} \right)}^2}}}{4}\]
Terms of \[4\sum\limits_{n = 1}^n {{n^2}} \] are \[{1^3} + {2^3} + {3^3} + \ldots \ldots + {n^3}\] and the sum is given as ${S_n} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$
And 1 + 2 + 3 + …. +n are the terms of \[\sum\limits_{n = 1}^n n \] whose sum is given as \[{S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\]
So the sum of equation 1 is \[{S_n} = \left( {3 \times \dfrac{{{{\left( {n\left( {n + 1} \right)} \right)}^2}}}{4}} \right) + \left( {4 \times \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) + \dfrac{{n\left( {n + 1} \right)}}{2}\]

\[{S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{3n\left( {n + 1} \right)}}{2} + \dfrac{{4\left( {2n + 1} \right)}}{3} + 1} \right)\]
$ \Rightarrow $\[{S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{3{n^2} + 3n}}{2} + \dfrac{{8n + 4}}{3} + 1} \right)\]
$ \Rightarrow $\[{S_n} = \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{9{n^2} + 9n + 16n + 8 + 6}}{6}} \right)\]
$ \Rightarrow $\[{S_n} = \dfrac{{n\left( {n + 1} \right)}}{{12}}\left[ {9{n^2} + 25n + 14} \right]\]
Hence \[\dfrac{{n\left( {n + 1} \right)}}{{12}}\left[ {9{n^2} + 25n + 14} \right]\] is the sum of the series 1.2.4 + 2.3.7 + 3.4.10 + …. To nth terms.

Note: The term series we introduced in the above question can be explained as the total of different numbers or elements that belong to a sequence or series as of now series are of two types one is infinite series and another is finite series. Infinite series consist of sums of infinite numbers belonging to a sequence whereas in finite series there are limited numbers of sums that belong to a sequence.