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Hint: This is not an AP or GP. It is a special series. Add 0 in the first term then subtract new series with the given series. You will find an AP is produced and move along with it and you will find the required solution.

__Complete step by step answer:__

Let

\[\begin{array}{l}

{S_n} = 5 + 11 + 19 + 29 + 41 + ...... + {a_{n - 1}} + {a_n}\\

{S_n} = 0 + 5 + 11 + 19 + 29 + 41 + ....... + {a_{n - 2}} + {a_{n - 1}} + {a_n}

\end{array}\]

Now subtract both of them we will get,

\[\begin{array}{l}

\Rightarrow {S_n} - {S_n} = 5 - 0 + [(11 - 5) + (19 - 11) + (29 - 19) + ....({a_{n - 1}} - {a_{n - 2}}) + ({a_n} - {a_{n - 1}})] - {a_n}\\

\Rightarrow 0 = 5 + [6 + 8 + 10 + 12 + .....{a_{n - 1}}] - {a_n}\\

\Rightarrow {a_n} = 5 + [6 + 8 + 10 + 12 + .....{(n - 1)^{th}}term]...............................(i)

\end{array}\]

Now it is clearly seen that \[6,8,10,12,.....{(n - 1)^{th}}term\]

We know that the general equation of an AP is given by \[{a_n} = a + (n - 1)d\]

Where a is the first term, n is the number of term and d is the common difference

Here in this case \[a = 6,d = 2\]

We know that sum of n terms of an AP is given by \[\dfrac{n}{2}\{ 2a + (n - 1)d\} \]

Now putting \[n = n - 1,a = 6\& d = 2\] we get

\[\begin{array}{l}

= \dfrac{{n - 1}}{2}[2(6) + \{ (n - 1) - 1\} 2]\\

= \dfrac{{n - 1}}{2}[12 + (n - 1 - 1)2]\\

= \dfrac{{n - 1}}{2}[12 + (n - 2)2]\\

= \dfrac{{n - 1}}{2}[12 + 2n - 4]\\

= \dfrac{{n - 1}}{2}[8 + 2n]\\

= \dfrac{{n - 1}}{2} \times 2[4 + n]\\

= (n - 1)(n + 4)

\end{array}\]

Now we can put this in equation (i) we get

\[\begin{array}{l}

\Rightarrow {a_n} = 5 + (n - 1)(n + 4)\\

\Rightarrow {a_n} = 5 + n(n + 4) - (n + 4)\\

\Rightarrow {a_n} = 5 + {n^2} + 4n - n - 4\\

\Rightarrow {a_n} = {n^2} + 3n + 1

\end{array}\]

Now,

\[\begin{array}{l}

{S_n} = \sum\limits_{n = 1}^n {{a_n}} \\

\Rightarrow {S_n} = \sum\limits_{n = 1}^n {{n^2} + 3n + 1} \\

\Rightarrow {S_n} = \sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n {3n} + \sum\limits_{n = 1}^n 1 \\

\Rightarrow {S_n} = \sum\limits_{n = 1}^n {{n^2}} + 3\sum\limits_{n = 1}^n n + \sum\limits_{n = 1}^n 1

\end{array}\]

We know that sum of all \[n \& {n^2}\] terms is given by \[\dfrac{{n(n + 1)}}{2}\& \dfrac{{n(n + 1)(2n + 1)}}{6}\] and also sum of n 1’s will naturally be n only.

\[\begin{array}{l}

\therefore \dfrac{{n(n + 1)(2n + 1)}}{6} + \left( {\dfrac{{n(n + 1)}}{2}} \right) + n\\

= \dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{3}{2}n(n + 1) + n\\

= \dfrac{{n(n + 1)(2n + 1) + 9(n + 1) + 6n}}{6}\\

= n\left( {\dfrac{{(n + 1)(2n + 1) + 9(n + 1) + 6}}{6}} \right)\\

= n\left( {\dfrac{{2{n^2} + 2n + n + 1 + 9n + 9 + 6}}{6}} \right)\\

= n\left( {\dfrac{{2{n^2} + 12n + 16}}{6}} \right)\\

= n\left( {\dfrac{{2({n^2} + 6n + 8)}}{6}} \right)\\

= \dfrac{n}{3}({n^2} + 6n + 8)\\

= \dfrac{n}{3}[n(n + 4) + 2(n + 4)]\\

= \dfrac{n}{3}[(n + 2)(n + 4)]\\

= \dfrac{{n(n + 2)(n + 4)}}{3}

\end{array}\]

Therefore the required sum is \[{\dfrac{{n(n + 2)(n + 4)}}{3}}\]

Note: The key step here was to find the AP from the given series and also remember the summation of \[n,{n^2}\& {n^3}\] which is \[\dfrac{{n(n + 1)}}{2},\dfrac{{n(n + 1)(2n + 1)}}{6}\& {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}\] respectively. These three are often used while trying to solve problems like this.

Let

\[\begin{array}{l}

{S_n} = 5 + 11 + 19 + 29 + 41 + ...... + {a_{n - 1}} + {a_n}\\

{S_n} = 0 + 5 + 11 + 19 + 29 + 41 + ....... + {a_{n - 2}} + {a_{n - 1}} + {a_n}

\end{array}\]

Now subtract both of them we will get,

\[\begin{array}{l}

\Rightarrow {S_n} - {S_n} = 5 - 0 + [(11 - 5) + (19 - 11) + (29 - 19) + ....({a_{n - 1}} - {a_{n - 2}}) + ({a_n} - {a_{n - 1}})] - {a_n}\\

\Rightarrow 0 = 5 + [6 + 8 + 10 + 12 + .....{a_{n - 1}}] - {a_n}\\

\Rightarrow {a_n} = 5 + [6 + 8 + 10 + 12 + .....{(n - 1)^{th}}term]...............................(i)

\end{array}\]

Now it is clearly seen that \[6,8,10,12,.....{(n - 1)^{th}}term\]

We know that the general equation of an AP is given by \[{a_n} = a + (n - 1)d\]

Where a is the first term, n is the number of term and d is the common difference

Here in this case \[a = 6,d = 2\]

We know that sum of n terms of an AP is given by \[\dfrac{n}{2}\{ 2a + (n - 1)d\} \]

Now putting \[n = n - 1,a = 6\& d = 2\] we get

\[\begin{array}{l}

= \dfrac{{n - 1}}{2}[2(6) + \{ (n - 1) - 1\} 2]\\

= \dfrac{{n - 1}}{2}[12 + (n - 1 - 1)2]\\

= \dfrac{{n - 1}}{2}[12 + (n - 2)2]\\

= \dfrac{{n - 1}}{2}[12 + 2n - 4]\\

= \dfrac{{n - 1}}{2}[8 + 2n]\\

= \dfrac{{n - 1}}{2} \times 2[4 + n]\\

= (n - 1)(n + 4)

\end{array}\]

Now we can put this in equation (i) we get

\[\begin{array}{l}

\Rightarrow {a_n} = 5 + (n - 1)(n + 4)\\

\Rightarrow {a_n} = 5 + n(n + 4) - (n + 4)\\

\Rightarrow {a_n} = 5 + {n^2} + 4n - n - 4\\

\Rightarrow {a_n} = {n^2} + 3n + 1

\end{array}\]

Now,

\[\begin{array}{l}

{S_n} = \sum\limits_{n = 1}^n {{a_n}} \\

\Rightarrow {S_n} = \sum\limits_{n = 1}^n {{n^2} + 3n + 1} \\

\Rightarrow {S_n} = \sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n {3n} + \sum\limits_{n = 1}^n 1 \\

\Rightarrow {S_n} = \sum\limits_{n = 1}^n {{n^2}} + 3\sum\limits_{n = 1}^n n + \sum\limits_{n = 1}^n 1

\end{array}\]

We know that sum of all \[n \& {n^2}\] terms is given by \[\dfrac{{n(n + 1)}}{2}\& \dfrac{{n(n + 1)(2n + 1)}}{6}\] and also sum of n 1’s will naturally be n only.

\[\begin{array}{l}

\therefore \dfrac{{n(n + 1)(2n + 1)}}{6} + \left( {\dfrac{{n(n + 1)}}{2}} \right) + n\\

= \dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{3}{2}n(n + 1) + n\\

= \dfrac{{n(n + 1)(2n + 1) + 9(n + 1) + 6n}}{6}\\

= n\left( {\dfrac{{(n + 1)(2n + 1) + 9(n + 1) + 6}}{6}} \right)\\

= n\left( {\dfrac{{2{n^2} + 2n + n + 1 + 9n + 9 + 6}}{6}} \right)\\

= n\left( {\dfrac{{2{n^2} + 12n + 16}}{6}} \right)\\

= n\left( {\dfrac{{2({n^2} + 6n + 8)}}{6}} \right)\\

= \dfrac{n}{3}({n^2} + 6n + 8)\\

= \dfrac{n}{3}[n(n + 4) + 2(n + 4)]\\

= \dfrac{n}{3}[(n + 2)(n + 4)]\\

= \dfrac{{n(n + 2)(n + 4)}}{3}

\end{array}\]

Therefore the required sum is \[{\dfrac{{n(n + 2)(n + 4)}}{3}}\]

Note: The key step here was to find the AP from the given series and also remember the summation of \[n,{n^2}\& {n^3}\] which is \[\dfrac{{n(n + 1)}}{2},\dfrac{{n(n + 1)(2n + 1)}}{6}\& {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}\] respectively. These three are often used while trying to solve problems like this.

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