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Thanks..its a joy to see these solutions! Regards

The cost of renting a bus for a field trip was divided evenly among 20 students.

At the last minute, 10 more students joined the trip.

The cost of renting the bus was then redistributed and evenly split among all of the students.

The cost for each of the original 20 students decreased by $1.50.

What was the total cost of renting the bus, in dollars?

\(\text{Let total cost $=c$} \\ \text{Let students $=s$} \\ \text{Let costs per student $c(s)=\dfrac{c}{s}$} \)

\(\begin{array}{|rcll|} \hline c(30) &=& c(20) - $1.50 \\\\ \dfrac{c}{30} &=& \dfrac{c}{20}- $1.50 \quad | \quad \cdot 30\\\\ c &=& 1.5c - $1.50\cdot 30 \\ c &=& 1.5c - $45 \\ 1.5c-c &=& $45 \\ 0.5c &=& $45 \\ c &=&\dfrac{ $45}{0.5} \\ \mathbf{c} &=& \mathbf{$90} \\ \hline \end{array}\)

The total cost of renting the bus, in dollars was $90

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

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

What was the total cost? How much did Vino have to pay?

Please answer these questions.

Without them, I can not solve the problem.

\(tommarvoloriddle \)

Nice, Pink Girl!

Sorry, CPhill, just wondering if you knew how to space in latex.

If you do not know, you would put a \ and a space at the end of each word in the latex section, hence,

has\ its\ foci\ at

which would equal to:

\(has\ its\ foci\ at\)

\(tommarvoloriddle \)

What is the sum of the following sequence:
\(\mathbf{1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000}\) .

Using the hockey stick identity and the  binomial coefficient  \(\dbinom nk\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ 1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000 } \\\\ &=& \dfrac{5!}{0!} + \dfrac{6!}{1!} + \dfrac{7!}{2!} + \dfrac{8!}{3!} + \dfrac{9!}{4!} + \dfrac{10!}{5!} + \dfrac{11!}{6!} +\ldots + \dfrac{1000!}{995!} \\\\ &=& \dfrac{5!}{5!0!}5! + \dfrac{6!}{5!1!}5! + \dfrac{7!}{5!2!}5! + \dfrac{8!}{5!3!}5! + \dfrac{9!}{5!4!}5! + \dfrac{10!}{5!5!}5! + \dfrac{11!}{5!6!}5! +\ldots + \dfrac{1000!}{5!995!}5! \\\\ &=& \dbinom 55 5! + \dbinom 65 5! + \dbinom 75 5! + \dbinom 85 5! + \dbinom 95 5! + \dbinom {10}5 5! + \dbinom {11}5 5! +\ldots + \dbinom {1000}5 5! \\\\ &=& 5! \left( \underbrace{\dbinom 55 + \dbinom 65 + \dbinom 75 + \dbinom 85 + \dbinom 95 + \dbinom {10}5 + \dbinom {11}5 +\ldots + \dbinom {1000}5}_{\text{long stick of the hockey in Pascals triangle}} \right) \\\\ &=& 5!*\left(\underbrace{\dbinom{1001}{6}}_{\text{bend of the hockey stick(hockey stick identity)} } \right) \\\\ &=& 5!* \dbinom{1001}{6} \\\\ &=& 5!* \dfrac{1001!}{6!995!} \\\\ &=& 120*1376423590241700 \\ &=& \mathbf{165170830829004000} \\ \hline \end{array}\)

The only reason why I guessed this was a hw homework problem was the extra latex, and it had dollar signs. Only AoPS and a few other math sites use that- and they have hw too...

And the DUE soon thing.

:) Thank you Melody.

Sorry for the confusion Tom.

I was not referring to you.

Thank you so much!

Thx so much!

Thx! so much

Thx!

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!!!

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

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

what? me? what did I do? or are you talking to the guest? I'm not sure...

Since you could not be bothered writing your question legibly, I wonder if you can be bothered copying CPhill's answer. 

But then maybe you only have to click and paste. It is not like you have to understand or write anything for yourself.

Maybe you can manage.

Wut? I think I am wrong... 

\(tommarvoloriddle\)

Great! maybe you just needed to wait...

Is this an AoPS HW problem? That's a no-no.

Working backwards....the last number displayed before pressing the  \(\textcolor{blue}{\bf\circledast}\) key would have been 324

Taking the square root of this we get 18

So...before pressing  the  \(\textcolor{blue}{\bf\circledast}\)    key   we would have had 81

Taking the square root of this  = 9  =  the original integer

Ginger is harsh Tom but she is also right.

You are offering answers to questions that you know nothing at all about.

You say people should find their own answers and I agree but showing them irrelevant websites is not helpful.

When you answer a question the asker has a fair expectation that you know something helpful. 

It can be incredibly frustrating, to an asker, when someone , who cannot answer, pretends that they can.

I really like your level of enthusiasm and I do not want to dampen it too much, or for too long. 

But please stay within your knowledge base.

Present your question properly.