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# Find the sum of the following series to n terms 1.4.7+4.7.10+7.10.13+....  Hint: In any series numbers follow a particular order. Before finding the sum of series, observe the given series to find that order to make simple calculations.

In the given series, the first term is 1.4.7, the second term is 4.7.10 and the third term is 7.10.13 and so on.

We have to find the general form of the series. Observing the given series,

 (3n-2) (3n+1) (3n+4) n=1 1 4 7 n=2 4 7 10 n=3 7 10 13

The given series follows an order, such that ‘n’ th term is ${T_n} = (3n - 2) \cdot (3n + 1) \cdot (3n + 4)$

$\Rightarrow {T_n} = 27{n^3} + 27{n^2} - 18n - 18$

Now we got a generalized form for the given series, we need to find the sum of the series.

Taking summation over ‘n’ on both sides of ${T_n}$

$\Rightarrow \sum {{T_n} = \sum {(27{n^3} + 27{n^2} - 18n - 8} } )$

$\Rightarrow \sum {{T_n} = \sum {(27{n^3}) + \sum {(27{n^2})} - \sum {(18n)} - \sum {(8)} } }$

$\Rightarrow \sum {{T_n} = 27\sum {({n^3}) + 27\sum {({n^2})} - 18\sum {(n)} - 8} } n$

$\left[ {\because \sum {{n^3}} = \dfrac{{{n^2}{{(n + 1)}^2}}}{4};\sum {{n^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6};\sum {n = \dfrac{{n(n + 1)}}{2}} ;\sum {k = n \times k} } \right]$

$\Rightarrow \sum {{T_n} = 27\left( {\dfrac{{{n^2}{{(n + 1)}^2}}}{4}} \right) + 27\left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right) - 18\left( {\dfrac{{n(n + 1)}}{2}} \right)} - 8n$

$\Rightarrow \sum {{T_n} = n(n + 1)\left[ {\dfrac{{27n(n + 1)}}{4} + \dfrac{{27n(2n + 1)}}{7} - \dfrac{{18}}{2}} \right]} - 8n$

$\Rightarrow \sum {{T_n} = n(n + 1)\left[ {\dfrac{{27({n^2} + n)}}{4} + \dfrac{{27(2{n^2} + n)}}{7} - 9} \right]} - 8n$

$\therefore$The sum of the given series is $\sum {{T_n} = n(n + 1)\left[ {\dfrac{{27({n^2} + n)}}{4} + \dfrac{{27(2{n^2} + n)}}{7} - 9} \right]} - 8n$

Note: First we found the general form for the given series. We used basic summation formulae to get the required solution. We applied those formulae and simplified the equation.
$\sum {k = n \times k}$, Where k is a constant.
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