[math]
{{2n}\choose{2}} = 4{{n}\choose{2}} + n
[/math]
>Veeky Forums can't prove this
[math]
>memebinatorics
...
Do your own homework faggot
wrong for n=1
Just did. See
It's not homework, I just stumbled upon it in my imaginings.
I remember I once proved doing something very ingenious.
Fuck induction, induction is for gay boys.
Compute [math] {{2n}\choose{2}} - 4{{n}\choose{2}} [/math]
First apply the definition of choose. Then merge both fractions into one and then use the definition of factorial to cancel a bunch of shit.
I was surprised, but you can actually reach [math] n [/math] if you are clever enough.
Good luck with your homework, Brainletto.
>manipulating symbols
>not double counting
[math]
\sum\limits_{k=1}^{n}k{{n}\choose{k}}=n2^{n-1}
[/math]
>Veeky Forums can't prove this
TUMBLING DOWN TUMBLING DOWN TUMBLING DOOOOOOOWWWWNNN
[math]1+1 = 2[/math]
>Veeky Forums can't prove this
>Then merge both fractions into one
The fractions are trivial to cancel.
[math]
\sum\limits_{i=0}^{k} {{m}\choose{i}} {{n}\choose{k-i}} = {{m+n}\choose{k}}
[/math]
>Veeky Forums can't even prove this
>Sci moves in, Sci moves out, you can't prove that
Get in the fucking robot u stupid emo kid
[eqn]\zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} [/eqn]
>Veeky Forums can't even prove this
But that's literally how it's defined
This is a definition, not a statement.
[math]\zeta(z) = 0 \implies \Re(z) = \frac{1}{2}[/math]
>Veeky Forums can't prove this
That just makes the proof short
Because it's not true
I'll prove it for you brainlet out there.
>Assume we are at a Veeky Forums meet up.
>For some reason only people from /pol/ and /tv/ showed up
> in fact there are only m+n people in total at the meetup: m from /pol/ and n from /tv/
> Question : how many sungrounded of i people can we make
> Answer 1: by definition , this is m+n choose i
>Answer 2: on the other hand , any i subgroup will have a certain number of k people from /pol/ and i-k from /tv/. As the k can range from 0 to I we sum and we are done...
[math]
\sum\limits_{i=0}^{k}{{n+i}\choose{i}}={{n+k+1}\choose{k}}
[/math]
Double count this you little shit.
In this case the proof from the book would be a very easy double counting argument for proving Pascal's identity (easy and classic), then trivially applying that by induction.