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Thank you so much for your help! Your solution is so clear and simple! I guess I should have thought of drawing PA & PB! Thanks!  

\(\sum\limits_{k=1}^{n}(k)(k+1)(k+2)\) .

The question implies that the real part of  \(\frac{1}{1-z}\)  is the same for all nonreal complex values of  z  such that  \(|z| = 1\) . So assuming that's true, we can pick any nonreal complex  z  such that  \(|z| = 1\)  and find the real part of  \(\frac{1}{1-z}\) .

Let's pick  z  =  0 + 1i

Then...

\({\frac{1}{1-z}}\ =\ \frac{1}{1-(0+1i)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\cdot\frac{1+1i}{1+1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-i^2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-(-1)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{2}+\frac{1}{2}i \)

The real part is  \(\boxed{\frac12}\)

We can check a few more cases to see that it is 1/2 for those as well:

We have these summations.....using the  sum of differences

3           11           26          50           85

        8          15           24          35

             7             9            11

                   2              2

                          0

We have  3 rows  of non-zero  differences  until we get to a zero difference row

This will  give us a 3rd degree polynomial   ax^3  + bx^2 + cx  + d

We  have the following  system

a   +  b   +  c  +  d  =   3

a(2)^3  + b(2)^2  + c(2)  + d  =  11

a(3)^3 + b(3)^2  + c(3)   + d  = 26

a(4)^3 + b(4)^2  + c(4) + d =  50

Simplifying this, we get

a + b + c + d  = 3

8a +  4b + 2c + d =  11

27a + 9b + 3c + d =  26

64a  + 16b + 4c + d  = 50

This system isn't difficult to solve, just tedious......I used WolframAlpha  to get that

a= 1/3   b = 3/2  c  = 7/6   d  = 0

So our polynomial is

f(n)  = (1/3)n^3  +  (3/2)n^2  + (7/6)

Agree with Hec        ....guest answer is incorrect

25 = 3x^2-3

0 = 3x^2-28      quadratic formula shows x = +- 2 sqrt21 / 3

Sub into g(x) :    (+2sqrt21/3)² + 2 sqrt 21/3 +1 = ...........

                and     (-2sqrt21/3)²  - 2sqrt21/3 +1 = .............    you can do the math from here...should get same as hecticar

Let's find what values of  x  make  f(x) = 25

f(x)  =  25

3x² - 3  =  25

3x²  =  28

x²  =  28/3

x   =  ±√[ 28/3 ]

Now let's plug in both of these into  g(f(x))

g( 25 )   =   g(f( √[ 28/3 ] )   =   ( √[ 28/3 ] )² + √[ 28/3 ] + 1   =   31/3 + √[ 28/3 ]

g( 25 )   =   g(f( -√[ 28/3 ] )   =   ( -√[ 28/3 ] )² - √[ 28/3 ] + 1   =   31/3 - √[ 28/3 ]

And adding these two together we get...

(  31/3 + √[ 28/3 ]  )   +   (  31/3 - √[ 28/3 ]  )   =   31/3 + 31/3   =   62/3

Working backwards

1372   was  3721   before she pressed the blue star key  the last time

Taking the square root of this produces    61

61   must have  been 16  before pressing  the blue star the first time

Taking the square root  of  16   = 

4 =  the original  integer

There are 4 creases in the 3x3   two vertical two horizontal

  18/4 = 4.5 inches per crease

     each stam is thus    4.5 in / 3 stamp = 1.5 in per stamp

In a 10 x 10 there are 18 creases

10 stamps x 1.5 in/stamp x 18 = 270 inches total

This is similar to  the toothpick problem

Number of perforations in the  3 x 3 sheet 

Number of   stamps in  each row  *  ( 3-1) rows  +    Number  of stamps in each column * ( 3 - 1) columns  =

(3) (3-1)  +  (3) ( 3-1)  =

3(2)   +  3 (2)  = 

12

So   each perforation  =   18 inches  / 12 =   1.5  in   

Using the  same reasoning  applied to  the  10 x 10  sheet

(10 (10 - 1)  +  10 (10 - 1)  =

10 (9)    +   10 (9)   = 

90  +  90   =   180  perforations

180 *  1.5   =

270 inches 

If the 3x3 and 10x10 square have the same proportions meaning they have same sized stamps and same sized creases, it's just a proportionality problem so take the area of the 10x10 over the 3x3 100/19. The multiply that by the length of the creases on the 3x3 sheet to get 200.

(Correct me if I'm wrong)

If we reflect the 30 60 90 triangle over the y axis, we get an equilateral triangle, congruent to the original one because they share a side. The side length is 12, as 6*2=12.

Now we have to find a formula for the area of an equilateral triangle.

If we draw an median to any side, we will have two congruent triangles, by SSS. Therefore the angles are the same and have to sum to 180, so the angles are 90. Therefore, a median is an altitude and vice versa with altitudes by AAS in an equilateral triangle. 

Let the side be x. Then, base is x/2 and height is sqrt (x^2 - x^2/4) = x sqrt 3 / 2. So, area of equilateral triangle = x^2 sqrt 3 /  4.

So, 144/4 = 36 so 36 sqrt 3.

woah thanks that was a lot

This problem sounds interesting:

If we draw PA and PB to make ∆PDA and ∆PCB, we have that the altitudes must be equal, so the areas of PAD and PBC are in the ratio 5:7.

If PX is the altitude to AD, we can treat AD as the base instead of PD, thus PX is the altitude and same for BC and PC. Thus, we have that PX*5/2 : PY*7/2 = 5:7, so 5PX : 7PY = 5:7. Therefore, dividing both sides by 5 and then multiplying by 5/7, we get PX : PY = 1:1, so PX = PY.

We have the form

x  = - a ( y - k)^2  +  h 

And the vertex  =  (5, -4) = (h, k)

x =  -a ( y - - 4)^2  + 5

x = -a  ( y + 4)^2  +   5

We know another point on the  graph  ( -3, 0)

So....we can solve for  a

-3  = -a ( 0 + 4)^2  + 5

-3 - 5 =  -a ( 16)

-8 = -a (16)

-8/-16 = a

1/2  = a

So we have

x = (-1/2) ( y + 4)^2  +  5     expand this

x= (-1/2) ( y^2  + 8y + 16)  + 5

x = (-1/2)y^2  -4y  -8 + 5

x = (-1/2)y^2 - 4y -3

a  +  b  +  c  =   (-1/2)  - 4 - 3  =   -7 1/2   =    - 15/2

Here's a graph :  https://www.desmos.com/calculator/fxczjujb5g

...then expand to    x = (-1/2)(y^2+8y+16) +5

                               x = -1/2y^2 -4y-3     take it from here   

Start with the vertex  5,-4

   vertex form x=   a (y+4)^2 +5     use point  (3,-2)   to calc  a =  - 1/2

x = -1/2 (y+4)^2 + 5

f(25) = 1872

g(f(25)) = 1872^2 + 1872+1 = 3506257

This is a permutation, as order matters.

Keep in mind that that permutation will only give you the possible, but you can do combinations when there is only one possible arrangement as it returns only one arrangment: 3 choose 3 = 1, whereas 3 pick 3 = 6

Favorable / possible

Possible would be 10 contestants, then choose 3 of them. 

There are 10 ways to choose first, 9 ways to choose second, 8 ways to choose 3rd.

720 possible.

For the favorable, we have 1 way to choose 1st, 1 way to choose 2nd, 1 way to choose 3rd.

So 1/720

\(501+503+\dots+697+699\\~\\ =\quad (501+2\cdot0) + (501+2\cdot1) +\dots+(501+2\cdot98) +(501+2\cdot99) \\~\\ =\quad (499+2\cdot1) + (499+2\cdot2) +\dots+(499+2\cdot99) +(499+2\cdot100) \\~\\ =\quad \sum\limits_{n=1}^{100}(499+2n) \\~\\ =\quad \sum\limits_{n=1}^{100}499+ \sum\limits_{n=1}^{100}2n \\~\\ =\quad 100\cdot499+ 2\cdot\sum\limits_{n=1}^{100}n \\~\\ =\quad 49900+ 2\cdot\frac{100(100+1)}{2} \\~\\ =\quad 49900+ 10100\\~\\ =\quad60000\).

We need to  solve this equation for  n  and then add 1 to the answer to findthe number of terms

501 + 2n  =699

2n =  699 - 501

2n = 198

n = 99

Number of terms  = 99 + 1   =100

Sum =   [ first term + last term ] [ number of terms / 2 ]

[ 501 + 699 ]  [ 100/2]  =

1200 * 50  =

60000

60000

The height should be decreased by 28%.

10! / (10-3)! = 10! / 7! = 720     One of which is correct order 1 - 2 - 3           =   1/720

that should be 1/6  x 1/120 = 1/720     jeez.....