[math] {{2n}\choose{2}} = 4{{n}\choose{2}} + n [/math] >/sci/

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>>8747792
>memebinatorics

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>>8747805

>>8747792
Do your own homework faggot

wrong for n=1

>>8747792
Just did. See >>8747818

>>8747827
It's not homework, I just stumbled upon it in my imaginings.

>>8747792
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.

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>>8747845
>manipulating symbols
>not double counting

[math]
\sum\limits_{k=1}^{n}k{{n}\choose{k}}=n2^{n-1}
[/math]
>/sci/ can't prove this

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>>8747792
TUMBLING DOWN TUMBLING DOWN TUMBLING DOOOOOOOWWWWNNN

[math]1+1 = 2[/math]
>/sci/ can't prove this

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>>8747845
>Then merge both fractions into one
The fractions are trivial to cancel.

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[math]
\sum\limits_{i=0}^{k} {{m}\choose{i}} {{n}\choose{k-i}} = {{m+n}\choose{k}}
[/math]
>/sci/ can't even prove this

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>>8747792
>Sci moves in, Sci moves out, you can't prove that

>>8747891
Get in the fucking robot u stupid emo kid

[eqn]\zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} [/eqn]

>/sci/ can't even prove this

>>8748238
But that's literally how it's defined

>>8748238
This is a definition, not a statement.

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[math]\zeta(z) = 0 \implies \Re(z) = \frac{1}{2}[/math]

>/sci/ can't prove this

>>8748254
>>8748257
That just makes the proof short

>>8748305
Because it's not true

>>8747920
I'll prove it for you brainlet out there.

>Assume we are at a 4chan 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...

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>>8748732
[math]
\sum\limits_{i=0}^{k}{{n+i}\choose{i}}={{n+k+1}\choose{k}}
[/math]
Double count this you little shit.

>>8748824
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.