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
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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\,

:D


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 bn.


 1 + \sum_{n=2}^4 n + 5 + 6 + 7 + 9 + 10 \neq 1 + \sum_{n=2}^4 (n + 5n) + 6 + 7 + 9 + 10