1.2.5 + 2.3.6 + 3.4.7 + ……..

Solution:

Let the nth term of the series here be given by T_n

We know that:

$ {{T}_{n}}~=\text{ }n\text{ }\left( n\text{ }+\text{ }1 \right)\text{ }\left( n\text{ }+\text{ }4 \right) $

$ =\text{ }n\text{ }\left( {{n}^{2}}~+\text{ }5n\text{ }+\text{ }4 \right) $

$ =\text{ }{{n}^{3}}~+\text{ }5{{n}^{2}}~+\text{ }4n $

Now, let the sum of n terms of the series be given by S_n

We have: