\[2 + 4 + 7 + 11 + 16 + ........n{\rm{ terms}}\]Equals
A. \[\dfrac{1}{6}\left( {{n^2} + 3n + 8} \right)\]
В. \[\dfrac{n}{6}\left( {{n^2} + 3n + 8} \right)\]
C. \[\dfrac{1}{6}\left( {{n^2} - 3n + 8} \right)\]
D. \[\dfrac{n}{6}\left( {{n^2} - 3n + 8} \right)\]

Hint: The series is \[2 + 4 + 7 + 11 + 16 + ........n{\rm{ terms}}\]therefore the first issue is figuring out its generalized form. Because there is no common difference or set common ratio in the given series, it cannot be solved using the AP and GP method. The provided series is neither an arithmetic progression nor a geometric progression. We shall solve them in accordance with the only suggestion they provided, which is five numbers in the form of addition.

Formula Used: The general formula of infinite series is
\[\dfrac{{n(n + 1)}}{2}\]

Complete step by step solution: We have been provided a series in the question.
Let us consider the series is
\[{\rm{S}} = 2 + 4 + 7 + 11 + 16 + \ldots \ldots n{\rm{ terms }}\]
Now, we have to write the above given series as below,
\[ \Rightarrow {\rm{S}} = \left( {1 + \dfrac{{{1^2} + 1}}{2}} \right) + \left( {1 + \dfrac{{{2^2} + 2}}{2}} \right) + \left( {1 + \dfrac{{{3^2} + 3}}{2}} \right) + \left( {1 + \dfrac{{{4^2} + 4}}{2}} \right) + 16 + \ldots \ldots \left( {1 + \dfrac{{{n^2} + n}}{2}} \right){\rm{ terms}}\]
Thus, in terms of formula for n terms, we can write the general expression as,
\[ \Rightarrow {a_n} = \left( {1 + \dfrac{{{n^2} + n}}{2}} \right)\]
Now, we have to determine the sum of the \[{{\rm{n}}^{{\rm{th}}}}\] term of the given series,
\[ \Rightarrow {{\rm{S}}_{\rm{n}}} = \sum\limits_{{\rm{n}} - 1}^\infty {{{\rm{a}}_{\rm{n}}}} \]
Now, substitute the general form of \[{a_n}\] to the above expression, we get
\[ \Rightarrow {{\rm{S}}_{\rm{n}}} = \sum\limits_{{\rm{n}} - 1}^\infty {\left( {1 + \dfrac{{{{\rm{n}}^2} + {\rm{n}}}}{2}} \right)} \]
And the required summation is as below,
\[ \Rightarrow {{\rm{S}}_{\rm{n}}} = \sum\limits_{{\rm{n}} - 1}^\infty 1 + \sum\limits_{{\rm{n}} - 1}^\infty {\dfrac{{{n^2}}}{2}} + \sum\limits_{{\rm{n}} - 1}^\infty {\dfrac{n}{2}} \]
Now, let us apply the formula we have,
\[\sum\limits_{{\rm{n}} - 1}^\infty 1 = n,\sum\limits_{{\rm{n}} - 1}^\infty {{n^2}} = \dfrac{1}{2}\left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right),\sum\limits_{{\rm{n}} - 1}^\infty n = \left( {\dfrac{{n(n + 1)}}{2}} \right)\]
\[ \Rightarrow {{\rm{S}}_{\rm{n}}} = n + \dfrac{1}{2}\left( {\dfrac{{n(n + 1)(2n + 1)}}{6}} \right) + \dfrac{1}{2}\left( {\dfrac{{n(n + 1)}}{2}} \right)\]
Now, we have to multiply the term in the denominator by \[\dfrac{1}{2}\] we get
\[ \Rightarrow {{\rm{S}}_{\rm{n}}} = n\left( {\dfrac{{12 + (n + 1)(2n + 1) + (3n + 3)}}{{12}}} \right)\]
Now, let’s simplify the numerator we get
\[ \Rightarrow {{\rm{S}}_{\rm{n}}} = n\left( {\dfrac{{12 + (n + 1)(2n + 4)}}{{12}}} \right)\]
Now, we have to take 2 as common, we have
\[ \Rightarrow {{\rm{S}}_{\rm{n}}} = n\left( {\dfrac{{6 + (n + 1)(n + 2)}}{6}} \right)\]
Now, we have to expand the above expression, we get
\[ \Rightarrow {{\rm{S}}_{\rm{n}}} = n\left( {\dfrac{{{n^2} + 3n + 2 + 6}}{6}} \right)\]
Now, we have to take 6 as common, we get
\[ \Rightarrow {{\rm{S}}_{\rm{n}}} = \dfrac{n}{6}\left( {{n^2} + 3n + 8} \right)\]
Therefore, \[2 + 4 + 7 + 11 + 16 + ........n{\rm{ terms}}\]equals \[{{\rm{S}}_{\rm{n}}} = \dfrac{n}{6}\left( {{n^2} + 3n + 8} \right)\]

Option ‘B’ is correct

Note: Students got confused mostly in these types of problems because the series given is neither arithmetic nor geometric. So, it is difficult to apply the formula, for that we have solved the series with the only given data a series in form addition to determine the solution using the general formula of infinite series of n terms.