1.2.5 + 2.3.6 + 3.4.7 + ……..
Solution:
Let the nth term of the series here be given by Tn
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 Sn
We have: