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Thank you Cphill and Electric Pavlov!

Mine might be more "math-y"......but it's also more " cook-book-y"

Your way required more ingenuity  [ IMHO  !!! ]

I think that the possible roots are 

 1 and -1 up to  4 times

2 and -2 once

3 an -3 up to twice     -  becasue 9 goes into 18

9 and -9  once

18 and -18 once.

So I think that the possible values of r are,      -3, -1, 1, 3

Thanks Guest that is a good answer.

I just counted them too but  I did not need to work out all the numbers.

I drew it in the shape of the normal triangle but insead of putting the actual number I just put  o  for odd and  e  for even.

If 2 odds are added the answer will be even

If 2 evens are added the answer will be even 

If one even and one odd are added the answer will be odd.

Just count them:

1   = R 0
1   1   = R 1
1   2   1   = R 2
1   3   3   1   = R 3
1   4   6   4   1   = R 4
1   5   10   10   5   1   = R 5
1   6   15   20   15   6   1   = R 6
1   7   21   35   35   21   7   1   = R 7
1   8   28   56   70   56   28   8   1   = R 8
1   9   36   84   126   126   84   36   9   1   = R 9
1   10   45   120   210   252   210   120   45   10   1   = R 10

I 'guess' this will work:

\( \sum_{1}^{20}\) (n-1)*7 - (n-2)    = 1160        Still rather clumsy....your answer is more elegant and 'math-y' 

EP's answer is more creative......but here's another way with the sum-of-differences

1         8            21            40  

      7        13             19

          6            6

We have two  non-zero rows....so.....this is a quadratic in the form  An^2 + Bn + C

So....we have this system of equations

A(1)^2  + B(1)  + C  = 1     ⇒  A + B + C  =  1    (1)

A(2)^2  + B(2) + C  = 8      ⇒ 4A + 2B + C  = 8   (2)

A(3)^2 + B(3)  +  C =  21   ⇒ 9A + 3B + C  = 21  (3)

Subtract (1) from (2)  and (2) from (3)   and we get

3A + B = 7     ⇒  -3A - B = -7   (4)

5A + B  = 13    (5)

Add  (4) and (5)  and we have that   2A  = 6    ⇒  A  = 3

And   3(3) + B = 7  ⇒  B = -2

And using (1)     3 - 2 + C  = 1  ⇒   C = 0

So....the generating polynomial  for the number of dots in the nth stellation   is    3n^2 - 2n

So....the number of dots in the 20th stellation is    3(20)^2 - 2(20)  = 1160

Which is the EXACT result that EP obtained !!!!

Crafty guesswork......trying to figure out a sigma notation for that mess-guess ! 

Very crafty, EP  !!!

A few things I noticed:

Other than that technical stuff I think it's really well written and easy to follow 👍

The final number in Row N  =  7N

So....the final number in Row 99  =  7(99)  = 693

So....the third number in Row 100  =  693 + 3  = 696

             1

+7-0      8

+2*7-1   21

+3*7 -2  40

+4*7-3   65

+5*7-4  96

+6*7 - 5  133

+7*7 -6  176

+8*7 - 7 225

+9*7 -8  280

+10*7 -9  341

+11*7 -10   408     ??

Oops.....I thought it said 12th

20th would be

12*7 -11

13*7 -12

14*7 -13

15*7-14

16*7 -15

17*7 -16

18*7 -17

19*7 - 18   =   1160    ??

Thank you!!!

I think we're getting off track, my point was that the proof made by max is incorrect, because the limit of the sequence a_n representing the integral is not one infinite series, but a sequence where each term of the sequence is a finite sum. Different  terms of the sequence a_n are sums that are made of different elements, as opposed to a series where a sequence of numbers gets added up one by one.

Max's argument boils down to "the limit of a_n is strictly positive because each member of a_n is strictly positive" but that argument is incorrect (for example: 1, 1/2, 1/3, 1/4....)

See below:

Yuck!!!....definitely didn't think on that one!!!!!

Let me re-work it  !!!!

If -5\leq a \leq -1$ and $1 \leq b \leq 3$, what is the least possible value of $\displaystyle\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right) $? Express your answer as a common fraction.

Cphill I don't think you can pick 2 twice, you have to pick two different balls. Your sum can never be 2 because to get 2 you have to pick ball number 1 twice.

Tim is doing a cakewalk with 11 steps. He takes his first step on step 1 and takes a total of 139 steps,
walking in a circle (so that after the 11th step he reaches the first step again).
Which step of the cakewalk does he end on?

I assume:

\(\begin{array}{|rcll|} \hline 139 &=& \underbrace{12\times 11}_{\text{position }1} + 7 \\ \text{Tim end on position } 1 + 7 = 8 \\ \hline \end{array}\)

Another way:

\(n(n+1)(n+2)(n+3)(n+4)=\frac{(n+4)!}{(n-1)!}=5!*\frac{(n+4)!}{(n-1)!*5!}={n+4\choose 5}*5! \)

"If \(-5\leq a \leq -1\) and \(1 \leq b \leq 3\) what is the least possible value of \(\displaystyle\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right)\)?"

The image below should help:

\(R_{\theta}=\begin{pmatrix}\cos(\theta)&-\sin(\theta)\\ \sin(\theta)&\cos(\theta)\end{pmatrix}\\~\\ \text{rotates a point counter-clockwise by angle $\theta$}\)

The volume of the prism is three times the volume of the pyramid.

The formula for the volume of a pyramid with height  h  and base area  B  is

V  =  \(\frac13\)Bh

x+y=6

y=6-x

Now (6-x) can be substituted for y

\(x^2+(-x+6)^2=28\)

Expanding this gets:

\(x^2+x^2-12x+36=28\)

\(2x^2-12x+8=0\)          Combine like terms and set equation to 0

\(x^2-6x+4=0\)              Divide by 2

This can't be factored so the quadratic equation needs to be used

\(x=\frac{6 \pm \sqrt{36-4(1)(4)}}{2(1)}\)

\(x=\frac{6 \pm \sqrt{20}}{2}\)

\(x=3+2\sqrt{5}\)

\(x=3-2\sqrt{5}\)

Plug these values into x+y=6 to get two points as the final answer.