Question

The sum of nth term of the series $1.2.5 + 2.3.6 + 3.4.7 + ......n$ terms is $\dfrac{1}{{12}}n\left( {n + 1} \right)\left( {3{n^2} + 23n + 34} \right)$
A) True
B) False

Answer 1

Hint:
$1.2.5 + 2.3.6 + 3.4.7 + ......n$. First, we have to find the general term. For this we can see $1.2.5 + 2.3.6 + 3.4.7 + ......n$ varies according to $n.\left( {n + 1} \right).\left( {n + 4} \right)$ hence general term is $n.\left( {n + 1} \right)\left( {n + 4} \right)$ on submission we know $\sum {n.\left( {n + 1} \right)\left( {n + 4} \right)} $ we know that $\sum {{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} $,\[\sum {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \]and $\sum {n = \dfrac{{n\left( {n + 1} \right)}}{2}}$

Complete step by step solution:
$1.2.5 + 2.3.6 + 3.4.7 + ......n$ terms
$1.\left( {1 + 1} \right).\left( {1 + 4} \right) + 2.\left( {2 + 1} \right).\left( {2 + 4} \right) + 3.\left( {3 + 1} \right).\left( {3 + 4} \right) + .......n$
Therefore, the general term
$ \Rightarrow n.\left( {n + 1} \right)\left( {n + 4} \right)$
$ \Rightarrow {n^3} + 5{n^2} + 4n$
Now, sum of the series will be
$\sum {n.\left( {n + 1} \right)\left( {n + 4} \right)} $
$\sum {{n^3} + 5{n^2} + 4n} $
$ \Rightarrow \sum {{n^3} + } 5\sum {{n^2} + 4} \sum n $
We know that $\sum {{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} $,\[\sum {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \]and $\sum {n = \dfrac{{n\left( {n + 1} \right)}}{2}} $
\[ \Rightarrow \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4} + \dfrac{{5n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \dfrac{{4n\left( {n + 1} \right)}}{2}\]
Taking common $n\left( {n + 1} \right)$
\[ \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{n\left( {n + 1} \right)}}{2} + \dfrac{{5\left( {2n + 1} \right)}}{3} + 4} \right)\]
$ \Rightarrow \dfrac{{n\left( {n + 1} \right)}}{2}\left( {\dfrac{{3{n^2} + 3n + 20n + 10 + 24}}{6}} \right)$
$ \Rightarrow \dfrac{1}{{12}}n\left( {n + 1} \right)\left( {3{n^2} + 23n + 34} \right)$

Note:
In these type of question 1st we find the general term then apply rule submission $\sum {{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} $,\[\sum {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \]and $\sum {n = \dfrac{{n\left( {n + 1} \right)}}{2}} $ for constant xn where x is constant.