User:PAStheLoD

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Hi there! I'm a 20-years-old net addict, at the moment. I was born in 1987, in Dunaújváros :) I've just decided to put up here something about myself .. if anyone is interested in me [I seriously doubt it ;], then feel free to ask, about [almost] anything :]

Wikipedia:Babel

hu	Ennek a szerkesztőnek magyar az anyanyelve.

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This user speaks English at a near-native level.

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Dieser Benutzer hat keine Deutschkenntnisse.

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7|-|15 |_|x0r 5|>34|<5 1337.

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This user reverts vandalism in the blink of an eye with Twinkle!

Here's the proof of a simple binomial coefficients related thingie.. I just needed Wikipedia's math rendering capability ;]
${n+1 \choose k}={n \choose k}+{n \choose k-1}$

$\frac{(n+1)!}{k!(n+1-k)!} = \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-(k-1))!}$

$\frac{(n+1)n!}{k!(n-k+1)(n-k)!} = \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-k+1)!}$

$\frac{(n+1)n!}{k!(n-k+1)(n-k)!} = \frac{n!}{k!(n-k)!} + \frac{n!k}{k!(n-k+1)!}$

$\frac{(n+1)n!}{k!(n-k+1)(n-k)!} = \frac{n!}{k!(n-k)!} + \frac{n!k}{k!(n-k+1)(n-k)!}$

$\frac{1}{k!(n-k)!} \frac{(n+1)n!}{n-k+1} =\frac{1}{k!(n-k)!} \left [ n! + \frac{n!k}{n-k+1} \right ]$

$\frac{(n+1)n!}{n-k+1} = n! + \frac{n!k}{n-k+1}$

$\frac{(n+1)}{n-k+1} = 1 + \frac{k}{n-k+1}$

$\frac{(n+1)}{n-k+1} = \frac{k+ (n-k+1)}{n-k+1}$

$n+1 =k+ (n-k+1) \,$

$n+1 = k+n-k+1\,$

$n+1 = n+1\,$

$f(x) \sim a_0 + \sum_{n=1}^\infty \left [ a_n \cos (nx) + b_n \sin (nx) \right ] \not\equiv f(x) \sim a_0 + \sum_{n=1}^\infty a_n \cos (nx) + b_n \sin (nx)$

$f(x) \sim a_0 + \sum_{n=1}^\infty a_n \cos (nx) + b_n \sin (nx) = f(x) \sim a_0 + \sum_{n=1}^\infty ( a_n \cos (nx) ) + b_n \sin (nx)$ so, here n is meaningless.. so as $b n$ .