Extremely confusing problem- show all you work and leave it in a expression plsssss

This pyramid has the following "closed form":

1/2[n^3 + n], where n = row number

1/2[2019^3 + 2019] =4,115,087,439

Look   at   the last number in each row  and notice this pattern

               1     =     sum of the 1st positive integer

                  3   =    sum of the first two positive integers

                     6 =    sum of the first three positive integers

                        10  =  sum of the first 4 positive integers

So.....the last entry in the nth row apears to be just the sum of the 1st  n positive integers

And the sum of the first n positive integers is just   :  n (n + 1)   /2

So....the last entry  in  the 2019th row is the sum of the first 2019 positive integers  = 

(2019) (2020)  / 2  =   2039190

And the nth row contains n integers...so....

We will have 2019  integers in this row and we need to find the first one....

Note that, after the first row....the first entry on any row is given by  :

Last entry  - row number + 1 

So...the first entry on the 2019th row must be :  2039190 - 2019 + 1  = 2037172

So...finally....the sum of the terms in the 2019th row is given by

[ first entry + last entry ] * number of terms / 2  =

[2037172 + 2039190 ] * [ 2019] / 2  = 

4,115,087,439

To generate a "closed" form solution that the Guest found......, we can use something known as the " sum of differences"

We have the following  sequence of row sums

1      5       15     34      65       111    take the positive difference between terms until we get a "constant" row

    4      10     19     31        46

         6       9      12     15

              3      3       3

It took us three  rows of differences to get to a "constant" row

This means that we will have a cubic (3rd power) "generating" polynomial  of the form

an^3  + bn^2  + cn + d = S(n)         where n is the row  number and S(n) is the row sum

So...we have this system of  equations 

a(1)^3 + b(1)^2 + c(1) + d  = 1

a(2)^3 + b(2)^2 + c(2) + d = 5

a(3)^3 + b(3)^2 + c(3) + d  = 15

a(4)^3  + b(4)^2 + c(4) + d  = 34   which gives us

a  +  b  +  c  + d   =   1

8a  + 4b + 2c + d  = 5

27a + 9b + 3c + d  = 15

64a + 16b + 4c + d  = 34

I'll  spare you the gory details of solving this by hand  [ but...I can go through it...if you want to ]

The solutions  are   a = 1/2, b = 0, c = 1/2   d = 0

So....the polynomial that generates the sum of any nth row is given by

(1/2)n^3 + (1/2)n  =  S(n)

[ n^3 +  n ]

________    =   S(n)

      2

Just as the "Guest" found!!!

What is the sum of the numbers in row 2019?

$\begin{array}{|lccccccc|} \hline \text{First row}: & & & &1 \\ \text{Second row}: & & &2 &, &3 \\ \text{Third Row}: & &4 &, &5 &, &6 \\ \text{Fourth row}: & 7 &, &8 &, &9 &, &10 \\ \hline \end{array}$

$\begin{array}{|l|ccccccccc|lcr|} \hline \text{row}& & & & & & & & & & &&\text{sum} \\ \hline 1 & & & & &1 & & & & & &=&1 \\ 2 & & & &\color{red}2 &, &\color{green}3 & & & & \left(\dfrac{2+3}{2}\right) \cdot 2 &=& 5 \\ 3 & & &\color{red}4 &, &5 &, &\color{green}6 & & & \left(\dfrac{4+6}{2}\right) \cdot 3 &=& 15 \\ 4 & & \color{red}7 &, &8 &, &9 &, &\color{green}10 & & \left(\dfrac{7+10}{2}\right) \cdot 4 &=& 34 \\ 5 & \color{red}11& , &12 &, &13 &, &14 &, &\color{green}15 & \left(\dfrac{11+15}{2}\right) \cdot 5 &=& 65 \\ \vdots & & & & &\vdots & & & & & \\ 2019 & & & & & a,\ldots ,b & & & & & \left(\dfrac{a+b}{2}\right) \cdot 2019 &=& \ ? \\ \hline n & & & & & a_n,\ldots ,b_n & & & & & \left(\dfrac{a_n+b_n}{2}\right) \cdot n &=& s_n \\ \hline \end{array}$

$\mathbf{a_n=\ ?}$
$\begin{array}{|lcccccccccc|} \hline \text{First row}: & d_1=1 & &2 & &4 & &7 & & 11 & \ldots \\ \text{Second Row}: & &d_2=1 & &2 & &3 & &4 & \ldots \\ \text{Third row}: & & &d_3=1 & &1 & & 1& \ldots \\ \hline \end{array}$

$\begin{array}{lcl} a_n &=& \dbinom{n-1}{0}\cdot d_1 + \dbinom{n-1}{1}\cdot d_2 + \dbinom{n-2}{2}\cdot d_3 \\\\ &=& \dbinom{n-1}{0}\cdot 1 + \dbinom{n-1}{1}\cdot 1 + \dbinom{n-2}{2}\cdot 1 \\\\ &=& 1 + (n-1)\cdot 1 + \dfrac{(n-2)(n-1)}{2\cdot 1}\cdot 1 \\\\ &=& 1 + n-1 + \dfrac{(n-2)(n-1)}{2} \\\\ \mathbf{a_n} &=& \mathbf{n +\dfrac{(n-2)(n-1)}{2}} \\ \end{array}$

$\mathbf{b_n=\ ?}$

$\begin{array}{|lcccccccccc|} \hline \text{First row}: & d_1=1 & &3 & &6 & &10 & & 15 & \ldots \\ \text{Second Row}: & &d_2=2 & &3 & &4 & &5 & \ldots \\ \text{Third row}: & & &d_3=1 & &1 & & 1& \ldots \\ \hline \end{array}$

$\begin{array}{lcl} b_n &=& \dbinom{n-1}{0}\cdot d_1 + \dbinom{n-1}{1}\cdot d_2 + \dbinom{n-2}{2}\cdot d_3 \\\\ &=& \dbinom{n-1}{0}\cdot 1 + \dbinom{n-1}{1}\cdot 2 + \dbinom{n-2}{2}\cdot 1 \\\\ &=& 1 + (n-1)\cdot 2 + \dfrac{(n-2)(n-1)}{2\cdot 1}\cdot 1 \\\\ &=& 1 + 2n-2 + \dfrac{(n-2)(n-1)}{2} \\\\ \mathbf{b_n} &=& \mathbf{2n-1 +\dfrac{(n-2)(n-1)}{2}} \\ \end{array}$

$\begin{array}{|rcll|} \hline s_n &=& \left(\dfrac{a_n+b_n}{2}\right) \cdot n \\\\ &=& \left(\dfrac{n +\dfrac{(n-2)(n-1)}{2}+2n-1 +\dfrac{(n-2)(n-1)}{2}}{2}\right) \cdot n \\\\ &=& \left(\dfrac{3n-1 + (n-2)(n-1)}{2} \right) \cdot n \\\\ &=& \left(\dfrac{3n-1 + n^2-3n+2}{2} \right) \cdot n \\\\ \mathbf{s_n} &=& \mathbf{\left(\dfrac{ n^2+1}{2} \right) \cdot n} \\ \hline \end{array}$

The sum of the numbers in row 2019

$\begin{array}{|rcll|} \hline \mathbf{s_n} &=& \mathbf{\left(\dfrac{ n^2+1}{2} \right) \cdot n} \\\\ s_{2019} &=& \left(\dfrac{ 2019^2+1}{2} \right) \cdot 2019 \\\\ s_{2019} &=& 2038181 \cdot 2019 \\\\ \mathbf{s_{2019}} &=& \mathbf{4115087439} \\ \hline \end{array}$