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Where does \(2^{99}\) come from?

Pascal's triangle:

See the sum of the each line:
\(\small{ \begin{array}{|l|rclcl|} \hline n \\ \hline 0 &\dbinom{0}{0} &=& 1 &=& 2^0 \\ 1 &\dbinom{1}{0}+\dbinom{1}{1} &=& 1+1 &=& 2^1 \\ 2 &\dbinom{2}{0}+\dbinom{2}{1}+\dbinom{2}{2} &=& 1+2+1 &=& 2^2 \\ 3 &\dbinom{3}{0}+\dbinom{3}{1}+\dbinom{3}{2}+\dbinom{3}{3} &=& 1+3+3+1 &=& 2^3 \\ 4 &\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} &=& 1+4+6+4+1 &=& 2^4 \\ 5 &\dbinom{5}{0}+\dbinom{5}{1}+\dbinom{5}{2}+\dbinom{5}{3}+\dbinom{5}{4}+\dbinom{5}{5} &=& 1+5+10+10+5+1 &=& 2^5 \\ \ldots \\ \color{red}99 & \dbinom{99}{0} + \dbinom{99}{1} + \dots + \dbinom{99}{98}+\dbinom{99}{99} &=&1+99+\dots + 99 + 1 &=& \color{red}2^{99} \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline && \color{red}(1+1)^n \\\\ &=&\sum \limits_{k=0}^{n}\dbinom{n}{k} \\ \\ &=& \dbinom{n}{0}+\dbinom{n}{1} + \dots + \dbinom{n}{n-1}+\dbinom{n}{n} \\\\ &=& \color{red}2^n \\ \hline \end{array}\)

like batman, there are many orphans such as harry potter and cinderella. All had original parents, but they died at a very young age or left them.

 what is wrong with you! I mean I have asked for help with homework, but I showed my efforts. If you tried, at least show it. I don't know whether you are struggling or not, but show that you at least tried. Otherwise You'll just be a lazy slob who expects others to do their work for him. Some people I know have tried this path and got expelled. Melody let you go easy, but seriously, block this website now. I use Blocksite, you should too. To be honest, if not the Web calc policy of no crude launguage, this would be much more graphical, don't be lazy, show some effort.  

P.S I did this on all your homework questions, if you do another homework question, I'll post this again, unless you show some effort. Just tell us what you did, even if it's completely off.

 What is wrong with you! I mean I have asked for help with homework, but I showed my efforts. If you tried, at least show it. I don't know whether you are struggling or not, but show that you at least tried. Otherwise You'll just be a lazy slob who expects others to do their work for him. Some people I know have tried this path and got expelled. Melody let you go easy, but seriously, block this website now. I use Blocksite, you should too. To be honest, if not the Web calc policy of no crude launguage, this would be much more graphical, don't be lazy, show some effort.  

Is this a homework problem? 

If not, don't continue reading this message.

If so, but you have not read Melody's message, please do so now.

If you have done both, then what is wrong with you! I mean I have asked for help with homework, but I showed my efforts. If you tried, at least show it. I don't know whether you are struggling or not, but show that you at least tried. Otherwise You'll just be a lazy slob who expects others to do their work for him. Some people I know have tried this path and got expelled. Melody let you go easy, but seriously, block this website now. I use Blocksite, you should too. To be honest, if not the Web calc policy of no crude launguage, this would be much more graphical, don't be lazy, show some effort.  

More civil wars would occur, children would be forced to join sides. Basically like Japan during shogun era.

Attn:  Yeager,

I just realized that almost all the guest questions are yours as well.

Behave yourself.  You will learn nothing by copying other people's answers.

hint:

\(\frac{54}{36}=?\\ \frac{24}{36}=?\\ \frac{16}{24}=?\)

Also, I have hidden most of your questions because we are not here to do all your homework for you.

Get an answer and learn from it. Then use what you have learned to answer other questions for yourself.

If you want to interact with me, or whoever is on the forum, over this question then you will probably learn a whole lot more and leave  a much better impression.

\(\sum_{k=1}^{6} -3(-4)^{k-1} = -3(-4)^{1-1} -3(-4)^{2-1}-3(-4)^{3-1}-3(-4)^{4-1}-3(-4)^{5-1}\)

                                 \(= -3(-4)^{0} -3(-4)^{1}-3(-4)^{2}-3(-4)^{3}-3(-4)^{4}\)

                                 \(= -3 + 12 - 48 +192 - 768\)

                                 \(= -615\)

let me know if I did anything wrong :)

That is correct. However, the question is asking in terms of n.

So, for example, if n = 1, then there are 2 options: 1 or 3

If n = 2, then there are 5 options: (11), (22), (33), (13), or (31)

We need to find some sort of recursion that works for any number n I think but I'm not sure how.

I have responded but it is not a proper answer.

Someone who is good at geometry should give this one a go.

I have not been able to work it out algebraically.

So I estimated it graphically. (using a free program GeoGebra)    [Not what you want, I know this]

The stupid pic has deleted itself so I cannot show it to you.  bummer!  

(Geogebra is playing up - this has not happened to me before)

The sides are approximately  in the ratio  65:62:59

The biggest angle is opposite the longest side.

Using cosine rule I got the biggest angle as approx 64.9 degrees.

Anyway. I am sure there is some  sensible algebraic way to do it.   

first let's find the 50th term. 

the "zeroth" term would be 16 - 3 = 13, meaning the 50th term would be 3(50) + 13 = 163. 

\(S_n = \frac{n(a_1 + a_n)}{2}\) where \(a_1\) is the first term and \(a_n\) is the last term of the sum you're finding.

\(S_{50} = \frac{50(16 + 163)}{2}\)

\(S_{50} = 25(179)\)

\(S_{50} = 4475\)

tell me if I did anything wrong :)

Also Batman isn't rea. Sorry, if that crushed your dreams.

You get birthed by your original parents. But something unfortunate happens and they decide to leave you at an orphanage because they can't take care of you, which is where you get adopted. 

For this we divide 24.60 by 410 to find the tax rate. 24.60/410 is 0.06. That's in decimal form. Multiply by 100 to get the percent form. 0.06 * 100 is 6. That's the number in percent form. The sales tax rate is 6%. 

$a_0 = 54 \div 3^3 = \frac{2}{3} = 2 \cdot 3^{-1}$

$a_1 = \frac{2}{3} \cdot 3 = 2 = 2 \cdot 3^0$

$a_2 = 2 \cdot 3 = 6 = 2 \cdot 3^1$

$a_3 = 6 \cdot 3 = 18 = 2 \cdot 3^2$

$a_4 = 18 \cdot 3 = 54 = 2 \cdot 3^3$

It should be clear by now that for $a_n$, it will be $\boxed{2 \cdot 3^{n-1}}.$

Nice. :))

=^._.^=

Yeah i realized taht a while ago lol. I got the same answer! pog

You're solving for geometric, not arithmetic. 

First, we need to look for the common ratio. 

x, x + y, x + 2y, x + 3y, x + 4y, ...

a_4 = x + 3y

a_16 = x + 15y

Subtract those 2 equations. 

125-41 = x + 15y - (x + 3y)

84 = 12y

y = 7 (that's the common ratio)

a_4 = x + 3y

41 = x + 3(7)

x = 20

So our first term is 20, then we add 7 for the next term. 

a_n = 20 + 7(n-1)

=^._.^=

i went to 125=41r^12 but now im stuck. pls help ;-;

wait so 41=a_1 r^3
ans 125=a_1  r^15

I was having trouble on how to find the common ratio. at first i was thinking as taht i needed to treat a_4 like a problem such as a_2=41 therefore 41=a_1*r but i can't figure out how to format it that way or if im even on the right path.