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Find the value of k so that
\(3 + \dfrac{3 + k}{4} + \dfrac{3 + 2k}{4^2} + \dfrac{3 + 3k}{4^3} + \dotsb = 8\).

\(\begin{array}{|rcll|} \hline 3*1 + (3 + k)\left(1*\frac{1}{4}\right) + (3 + 2k)\left(1* \frac{1}{4^2} \right) + (3 + 3k)\left(1* \frac{1}{4^3} \right) + \dotsb + (3 + (n-1)k)\left(1*\left(\frac{1}{4}\right)^{n-1}\right) + \dotsb = 8 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{arithmetical sequence:} \\ {\color{red}3}+({\color{red}3}+k)+({\color{red}3}+2k)+({\color{red}3}+3k)+\dotsb \\ \boxed{a={\color{red}3},\ k \text{ is the common difference} } \\\\ \text{geometric series:} \\ {\color{blue}1}+{\color{blue}1}*{\color{green}\frac{1}{4}}+{\color{blue}1}*({\color{green}\frac{1}{4}})^2+{\color{blue}1}*({\color{green}\frac{1}{4}})^3+\dotsb \\ \boxed{ b={\color{blue}1},\ r={\color{green}\frac{1}{4}}\ \text{ is the common ratio} } \\ \hline \end{array}\)

The sum  of the first n terms of an arithmetico–geometric sequence has the form:

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &=& \mathbf{ \dfrac{ ab-(a+nk)br^n} {1-r} + \dfrac{kbr(1-r^n)} {(1-r)^2}} \\ &&\boxed{a=3,\ b=1,\ r=\frac{1}{4}} \\\\ &=& \dfrac{3*1-(3+nk)*1*\frac{1}{4^n}}{1-\frac{1}{4}} + \frac{ k*1*\frac{1}{4}(1-\frac{1}{4^n})} { (1-\frac{1}{4})^2 } \\\\ &=&\frac{4}{3}\left( 3-\frac{3}{4^n}-\frac{nk}{4^n} \right) +\frac{16}{9}\left( \frac{k}{4}-\frac{k}{4*4^n} \right) \\\\ &=& 4-\frac{1}{4^{n-1}}-\frac{nk}{3*4^{n-1}} + \frac{4}{9}k-\frac{k}{4*4^n}*\frac{4*4}{9} \\\\ &=& 4-\frac{1}{4^{n-1}}-\frac{nk}{3*4^{n-1}} + \frac{4}{9}k-\frac{k}{9*4^{n-1}} \\\\ &=& 4+ \frac{4}{9}k -\frac{1}{4^{n-1}}\left(1+\frac{nk}{3}+\frac{k}{9}\right) \\ &&\boxed{\lim \limits_{n\to\infty} \frac{1}{4^{n-1}}\left(1+\frac{nk}{3}+\frac{k}{9}\right)=0} \\ 8 &=& 4+ \frac{4}{9}k \\ 4 &=& \frac{4}{9}k \quad | \quad : 4 \\ 1 &=& \frac{1}{9}k \quad | \quad * 9 \\ 9 &=& k \\ \mathbf{k} &=& \mathbf{9} \\ \hline \end{array}\)

Formula:

infinite arithmetico–geometric sequence

\(\begin{array}{|rcll|} \hline \mathbf{s} &=& \mathbf{\left(\dfrac{b}{1-r}\right) \Bigg( a+k\left( \dfrac{r}{1-r} \right) \Bigg)} \\ \hline \end{array}\)

\(p = \dfrac{\dbinom{4}{2}\dbinom{11}{1}+\dbinom{5}{2}\dbinom{10}{1}+\dbinom{6}{2}\dbinom{9}{1}}{\dbinom{15}{3}}\)

I like your drawings Dragan    

What is the largest integer n such that 3^n is a factor of 1 x 3 x 5 x... x 97 x 99?

How many of those numbers has a factor of 3

3,6,9, ........99     There are 33 of them

Divide each of these by 3

1,2,3,......  33

Which of these have a factor of 3

3,6,9, ... 33       There are 11 of them

Divide each of these by 3

1,2,3,......  11

Which of these have a factor of 3

3,6,9      There are 3 of them

Divide each of these by 3

1,2,3

Which of these have a factor of 3

3      There are 1 of them

33+11+3+1= 48

I think that the largest n is 48

AB and CD are two parallel chords of a circle such that AB = 24 cm and CD= 10 cm. If the radius of the circle is 13 cm , find the distance between the chords.

The distance between 2 chords is 17 cm

The interior angles of a convex nonagon have degree measures that are integers in an arithmetic sequence. What is the smallest angle possible in this nonagon?

Can you do any of this for yourself?

For example.  

What is a nonagon?

What is the angle sum of a nonagon?

What is the formula for an arithmetic sum?

Answer these and then I will help more.

No other answerer interfere, thank you.  

NO = 5/(1+sqrt(2)) = 2.071068294

MO = r / sin(45) = 2.92893287

MN = MO - NO =0.857864576

Proportion:           MN  :  NO  =  MT  :  TR

Radius of circle "M" is:      TR = 12.071068 m   

A bag contains 4 red marbles, 5 yellow marbles, and 6 blue marbles. Three marbles are to be picked out randomly (without replacement). What is the probability that exactly two of them have the same color?

15 marbles.  Let each one have a unique number, so in this sense, there are all different.

How many combinations are there

All red    4C3=4

All yellow    5C3=10

All blue   6C3=20

All the same colour=34

Exactly 2 red = 4C2*11 = 66

Exactly 2 yellow = 5C2*10=100

Exactly 2 blue = 6C2*9=135

Exactly 2 the same colour = 301

All different = 4*5*6=120

So

P(exactly two of them have the same color) = 301 /(23+301+120) = 301 / 444

(This is my logic, not necessarily correct)

By arithmetico-geometric series, 

\(a_1 g_1 + (a_1 + d)(g_1 r) + (a_1 + 2d)(g_1 r^2) + \dots = \frac{dg_2}{(1 - r)^2} + \frac{x_1}{1 - r}\)

(see https://artofproblemsolving.com/wiki/index.php/Arithmetico-geometric_series)

This gives an answer of k = 7.

Please put only one question per post.

AND leave out the meaningless $ signs.

To me this looks like the post of a very lazy person who cannot even be bothered presenting his question properly.

Yes, sorry, my mistake.

Yes that is right.

We can use Legendre's formula: https://en.wikipedia.org/wiki/Legendre%27s_formula

By Legendre's formula, the number of factors of 3 is

\(\left\lfloor \frac{99}{3} \right\rfloor + \left\lfloor \frac{99}{3^2} \right\rfloor + \left\lfloor \frac{99}{3^3} \right\rfloor + \left\lfloor \frac{99}{3^4} \right\rfloor = 33 + 11 + 3 + 1 = 48\)

Thank geno! I got 18/343

Probability of being used is 0.39.

Probability of not being used is 0.61.

You want exactly 7 of 9.

₉C₇ · 0.39⁷ · 0.61²         =         your answer

If you buy 7 hamburgers and 3 slices of pizza from the Sizzler, you get Sl in change back from a
$20 bill. If you buy 8 hamburgers and 2 slices of pizza, you get $0.50 back from a $20 bill. Find
the price of a hamburger and the price of a slice of pizza.  

See the part I highlighted in blue?  Sl doesn't make sense, so I'm going to assume you meant $1. 

Let H stand for the price in $ of a hamburger, and

Let P stand for the price in $ of a slice of pizza. 

So                                    7H + 3P  =  20 – 1       =  19.00 

and                                  8H + 2P  =  20 – 0.50  =  19.50 

Multiply both sides of the 1st equation by 2 and get       14H + 6P  =  38.00 

Multiply both sides of the 2nd equation by 3 and get      24H + 6P  =  58.50 

Subtract the fourth equation from the third and get       –10H  =  –20.50  

Divide both sides of this equation by –10 and get               H  =  2.05 

Plug 2.05 into the first equation in place of H                   (7)•(2.05) + 3P  =  19.00 

                                                                                                            3P  =  19.00 – 14.35  =  4.65 

                                                                                                              P  =  1.55 

_.

The probability of being born on a Tuesday is 1/7.

The probability of not being born on a Tuesday is 6/7.

You want to choose 2 out of a possible 3.

₃C₂ · (1/7)² · (6/7)¹  =  your answer

On the link, it was 121. But okay.

2/7?

It's the probability they are born on a Tuesday versus another day

What geno said.