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Can we assume that  4R is meant to be in brackets?

"Boy b***h"?  No, rather, "insufferable b***h"??

Actually, Tommarvoloriddle, I thought you were transgender....  Sorry! 

My name is Ginger Alexandra, and it annoys me too, when someone thinks I’m a boy. No one ever thinks that when seeing me in person, but on line, even using my name, Ginger, it’s rather common in academic circles (not in social circles). I suppose it’s mainly because of my logical, active-voice, writing style.  I can see that my acerbic commentaries and trolling might compel the perception of a bitch, and sometimes a “boy bitch.”

Euler's Theorem gives us 9^2004phi100=9^4  mod 100=3^8 mod 100=6561 mod 100= 61 mod 100, and 6+1=7.

What is the value of the sum$\dfrac {1}{1\cdot 3} + \dfrac {1}{3\cdot 5} + \dfrac {1}{5\cdot 7} + \dfrac {1}{7\cdot 9} + \cdots + \dfrac {1}{199\cdot 201}$?

Express your answer as a fraction in simplest form.

$\begin{array}{rcll} && \dfrac{1}{1*3} + \dfrac{1}{3*5} + \dfrac{1}{5*7}+ \dfrac{1}{7*9}+\ldots+\dfrac{1}{199*201} \\ &=& \dfrac{1}{1*3} + \dfrac{1}{3*5} + \dfrac{1}{5*7}+ \dfrac{1}{7*9}+\ldots+\dfrac{1}{(2n-1)(2n+1)} \\ \hline && \dfrac{1}{(2n-1)(2n+1)} = \dfrac12\left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right) \\ && \dfrac{1}{1*3} = \dfrac12\left( \dfrac{1}{1} - \dfrac{1}{3} \right) \\ && \dfrac{1}{3*5} = \dfrac12\left( \dfrac{1}{3} - \dfrac{1}{5} \right) \\ && \dfrac{1}{5*7} = \dfrac12\left( \dfrac{1}{5} - \dfrac{1}{7} \right) \\ && \dfrac{1}{7*9} = \dfrac12\left( \dfrac{1}{7} - \dfrac{1}{9} \right) \\ && \ldots \\ && \dfrac{1}{199*201} = \dfrac12\left( \dfrac{1}{199} - \dfrac{1}{201} \right) \\ \hline &=& \dfrac12\left( \dfrac{1}{1} - \dfrac{1}{3} \right) + \dfrac12\left( \dfrac{1}{3} - \dfrac{1}{5} \right) + \dfrac12\left( \dfrac{1}{5} - \dfrac{1}{7} \right) + \dfrac12\left( \dfrac{1}{7} - \dfrac{1}{9} \right)+\ldots+\dfrac12\left( \dfrac{1}{199} - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( \dfrac{1}{1} - \underbrace{\dfrac{1}{3} + \dfrac{1}{3}}_{=0} - \underbrace{\dfrac{1}{5}+\dfrac{1}{5}}_{=0} - \underbrace{\dfrac{1}{7} + \dfrac{1}{7}}_{=0} - \underbrace{\dfrac{1}{9}+ \dfrac{1}{9}}_{=0} +\ldots- \underbrace{\dfrac{1}{199}+\dfrac{1}{199}}_{=0} - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( \dfrac{1}{1} - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( 1 - \dfrac{1}{201} \right) \\ &=& \dfrac12\left( \dfrac{201-1}{201} \right) \\ &=& \dfrac12\left( \dfrac{200}{201} \right) \\ &=& \dfrac{100}{201} \\ \end{array}$

Use the hockey stick identity to solve this question

$\text{The formula seems to be }\\ \sum \limits_{k=1}^n \dfrac{1}{(2k-1)(2k+1)} = \dfrac{n}{2n+1}$

Let's see if we can prove this by induction

$P_1:\dfrac{1}{1\cdot 3} = \dfrac{1}{2(1)+1} = True\\ \text{Assume $P_n$ and prove $P_n \Rightarrow P_{n+1}$}\\~\\ \text{Let $S_n = \sum \limits_{k=1}^n \dfrac{1}{(2k-1)(2k+1)}$}\\ S_{n+1} = S_n + \dfrac{1}{(2n+1)(2n+3)} =\\ \dfrac{n}{2n+1}+\dfrac{1}{(2n+1)(2n+3)} = \\ \dfrac{1}{2n+1}\left(n + \dfrac{1}{2n+3}\right) = \\$

$\dfrac{1}{2n+1}\cdot \dfrac{2n^2+3n+1}{2n+3} = \\ \dfrac{n+1}{2n+3} = \dfrac{n+1}{2(n+1)+1}\\ \text{and thus $P_n \Rightarrow P_{n+1}$}$

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

$\begin{array}{|rcll|} \hline &&\mathbf{ 1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000 } \\ &=& \sum \limits_{k=1}^{996} k(k+1)(k+2)(k+3)(k+4) \\ &=& \sum \limits_{k=1}^{996} k^5 + 10 k^4 + 35 k^3 + 50 k^2 + 24 k \\ &=& \sum \limits_{k=1}^{996} k^5 + 10\sum \limits_{k=1}^{996} k^4 + 35 \sum \limits_{k=1}^{996}k^3 + 50\sum \limits_{k=1}^{996} k^2 + 24\sum \limits_{k=1}^{996} k \\ && \boxed{ \sum \limits_{k=1}^{n} k^1= \dfrac12 n^2 + \dfrac12 n^1 \\ \sum \limits_{k=1}^{n} k^2= \dfrac13 n^3 + \dfrac12 n^2 + \dfrac16 n^1 \\ \sum \limits_{k=1}^{n} k^3= \dfrac14 n^4 + \dfrac12 n^3 + \dfrac14 n^2 \\ \sum \limits_{k=1}^{n} k^4= \dfrac15 n^5 + \dfrac12 n^4 + \dfrac13 n^3 -\dfrac1{30} n^1 \\ \sum \limits_{k=1}^{n} k^5= \dfrac16 n^6 + \dfrac12 n^5 + \dfrac{5}{12} n^4 -\dfrac1{12} n^2 } \\ &=& \dfrac16 996^6 + \dfrac12 996^5 + \dfrac{5}{12} 996^4 -\dfrac1{12} 996^2 + 10\left(\dfrac15 996^5 + \dfrac12 996^4 + \dfrac13 996^3 -\dfrac1{30} 996 \right) \\ &&+ 35\left( \dfrac14 996^4 + \dfrac12 996^3 + \dfrac14 996^2 \right) + 50\left( \dfrac13 996^3 + \dfrac12 996^2 + \dfrac16 996 \right) \\ &&+ 24\left( \dfrac12 996^2 + \dfrac12 996 \right) \\ &=& \dfrac16 996^6 \\ && + \dfrac12 996^5 + \dfrac{10}5 996^5 \\ && + \dfrac{50}{3} 996^4+ \dfrac{10}2 996^4+ \dfrac{35}4 996^4 \\ && + \dfrac{10}3 996^3+ \dfrac{35}2 996^3+ \dfrac{50}3 996^3 \\ && -\dfrac1{12} 996^2 + \dfrac{35}4 996^2 + \dfrac{50}2 996^2 + \dfrac{24}2 996^2 \\ && -\dfrac{10}{30} 996 + \dfrac{50}6 996 + \dfrac{24}2 996 \\ &=& \dfrac16 996^6 + \dfrac{5}{2}\cdot 996^5 + \dfrac{85}{6} 996^4 + \dfrac{75}2 996^3 + \dfrac{274}6 996^2 +20\cdot 996 \\ &=& \mathbf{165170830829004000} \\ \hline \end{array}$

Connection reset probably means that the wifi is flickering on and off. Mine does too, it's spectrum...

😫😩😢😭😰😥😓

Hm how to check if i didn't change anything in Internet setting?(Not chrome but internet in general) 

Me too, I didn't pay was just checking it if it is the same or not 

These websites says: (The same all)

:

This site can't be reached The connection was reset. what does that mean?

Try:

checking the connection
checking the proxy server

Running windows network diagnostics 

Oh, nice! Unfortunately, you have to pay... So I was personally never interested in that website.

It works on my laptop just fine too. 

Another website: https://www.coursera.org/

I know, I said, "CONFIRMED" so I basically took a look at yours, said, "seems legit" checked mine, made a smiley face emoji: :-)

If you look at the work I presented to you, you would see that $\frac{6-9i}{13}$ is the answer I got I just simplified it as $\frac{6}{13}-\frac{9i}{13}$ which are basically the same thing. 

But your welcome, happy to help ;P

Thank you, Emerald Wonder (pink girl)! I confirmed my answer with yours- and I got $\frac{6-9i}{13}$

I think that is close enough.

PEACE!

$tommarvoloriddle$

$\frac{3}{2+3i}$

To remove $2+3i$ from the denominator, multiply the numerator and denominator by its complex conjugate

$\frac{3(2-3i)}{(2+3i)(2-3i)}$

Solve the denominator seperately

$(2+3i)(2-3i)$

${2}^{2}+{3}^{2}$

$13$

Now you have

$\frac{3(2-3i)}{13}$

Expand $3(2-3i)$ by distributing

$3(2-3i)$

$6-9i$

Now you would have

$\frac{6-9i}{13}=\frac{6}{13}-\frac{9}{13}i$

Your answer would be

$\frac{6}{13}-\frac{9}{13}i$

Hope this Helps ;P

Given that m and n are positive integers such that m is congruent to 6 (mod 9) and n is congruent to 0 (mod 9), 

what is the largest integer that mn is necessarily divisible by?

I assume,

$\begin{array}{|rclcl|} \hline m &\equiv& 6 \pmod {9} & \text{or} & m=6+9u,\quad u \in \mathbb{Z} \\ n &\equiv& 0 \pmod {9} & \text{or} & n=9v,\quad v \in \mathbb{Z} \\\\ mn &=& 9v(6+9u) \\ &=& 54v + 81uv \quad &|\quad gcd(54,81)=27 \\ \hline \end{array}$

If m and n are positive integers, then the largest integer that mn is necessarily divisible by is 27

What other websites?

nope , not yet.. :( If anyone had something similar let me know..

It isn't just khan academy website

I have 2-3 other websites which says the same.. Maybe they will work after few hours. 

Ginger, maybe she’s a “Tom....Boy” 

Is it working?