Given the sequence \[2,6,12,20,....\]. What is the formula for the \[{n^{th}}\] term and the sum of \[n\] terms?

Hint:Here, we use the concepts of patterns and their properties to solve this problem. We will also come across some standard results like sum of n natural numbers is equal to \[\dfrac{{n(n + 1)}}{2}\] and sum of squares of $n$ natural numbers is equal to \[\dfrac{{n(n + 1)(2n + 1)}}{6}\], while solving this problem.

Complete step by step answer:
The sequence of numbers given here are \[2,6,12,20,....\]So, if you observe in the sequence, the difference between the first two terms is 4. The difference between the next two terms is 6. And the difference between the next two terms is 8.So, this follows a pattern.Now we will rewrite the sequence as
\[(2),(2 + 4),(2 + 4 + 6),(2 + 4 + 6 + 8),.......\]
Taking 2 common from each term, we get,
\[2(1),2(1 + 2),2(1 + 2 + 3),2(1 + 2 + 3 + 4),.......\]
So, we can conclude that, \[{n^{th}}\] term is equal to two times the sum of \[n\] natural numbers.So,
\[{n^{th}}term = 2(1 + 2 + 3 + ....... + n)\]
\[ \Rightarrow {n^{th}}term = 2\left( {\dfrac{{n(n + 1)}}{2}} \right)\] -----as we all know that, sum of \[n\] natural numbers is \[\dfrac{{n(n + 1)}}{2}\]
\[ \Rightarrow {n^{th}}term = n(n + 1)\]
So, \[{n^{th}}term = {n^2} + n\]

Now, let us add these numbers and we get,
\[(2) + (2 + 4) + (2 + 4 + 6) + (2 + 4 + 6 + 8) + ....... + ({n^2} + n)\]
We can simply write this as
\[(2) + (2 + 4) + (2 + 4 + 6) + (2 + 4 + 6 + 8) + ....... + ({n^2} + n) = \sum\limits_{i = 1}^n {({i^2} + i)} \]
So, sum of \[n\] terms is given by
\[{S_n} = \sum\limits_{i = 1}^n {({i^2} + i)} \]
\[ \Rightarrow {S_n} = \sum\limits_{i = 1}^n {{i^2} + \sum\limits_{i = 1}^n i } \] -----(1)
Here, the first term i.e., \[\sum\limits_{i = 1}^n {{i^2}} \] represents the sum of squares of \[n\] natural numbers. And we all know that the sum of squares of \[n\] natural numbers is \[{1^2} + {2^2} + {3^2} + ........ + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}\]
So, \[\sum\limits_{i = 1}^n {{i^2}} = \dfrac{{n(n + 1)(2n + 1)}}{6}\]
And the second term i.e., \[\sum\limits_{i = 1}^n i \] represents the sum of \[n\] natural numbers. And sum of \[n\] natural numbers is \[1 + 2 + 3 + 4 + ....... + n = \dfrac{{n(n + 1)}}{2}\]

So, \[\sum\limits_{i = 1}^n i = \dfrac{{n(n + 1)}}{2}\]
So, upon substituting these values in (1),
\[ \Rightarrow {S_n} = \dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{{n(n + 1)}}{2}\]
\[ \Rightarrow {S_n} = \dfrac{{n(n + 1)}}{2}\left( {\dfrac{{2n + 1}}{3} + 1} \right)\]
On simplifying the equation,
\[ \Rightarrow {S_n} = \dfrac{{n(n + 1)}}{2}\left( {\dfrac{{2n + 4}}{3}} \right)\]
\[ \Rightarrow {S_n} = \dfrac{{n(n + 1)}}{2}(2)\left( {\dfrac{{n + 2}}{3}} \right)\]
So, finally we conclude with
\[\therefore {S_n} = \dfrac{{n(n + 1)(n + 2)}}{3}\]
And this is the sum of \[n\] terms of the given sequence.

Hence, the formula for the \[{n^{th}}\] term and the sum of \[n\] terms are \[{n^2} + n\] and \[\dfrac{{n(n + 1)(n + 2)}}{3}\].

Note:We used sigma functions here, to solve this problem and this function is distributive. That means, \[\sum\limits_{i = a}^b {f(i) + g(i) = \sum\limits_{i = a}^b {f(i) + \sum\limits_{i = a}^b {g(i)} } } \]
And also, be careful while simplifying the equations. And also, check your answer by substituting n values as 1,2 and 3 (or any other numbers).