This is true?

this is true?

yes

what the fuck? this is elementary shit

no

maybe

it takes less than one minute to prove this is true
you jackass

when n=1 only

bottom row, n+n=n+1
therefore n=1 is the only solution. Check that it works for the top row, and 1+(1+1)=1+2

this is bait right? sometimes I can't tell with you retards

This is the only correct answer.

I think you meant to multiply by the transpose of the second matrix

ok, and this?

Yeah, cuz
[math]C(n,n)+C(n+1,n)=\frac{n!}{n!(n-n)!}+\frac{(n+1)!}{n!(n+1-n)!}=n+2=\frac{(n+2)!}{(n+1)!(n+2-n-1)!}=C(n+2,n+1)[/math]

also kys

n=1 cuz n+n = n+1 gives n=1

k=0 cuz k+k+1 = k+1 gives k=0

also stop jacking off, and learn your arithmetic tables

how do you do this?
i know latex, but how do you do it in a post?

\begin{equation}
\alpha+\beta = \gamma
\end{equation}

[latex]\sigma[latex]

You put it in [math][/math ] tags. You can click the TeX button on the top-left of the reply form to test it.

[math]\sigma+\beta = \gamma^{\alpha}[/math]

[eqn]
A = U \Sigma V^{T} =\sigma_{1}\textbf u_{1} \textbf v_{1}^{T}+\sigma_{2}\textbf u_{2} \textbf v_{2}^{T}+\cdots +\sigma_{n}\textbf u_{n} \textbf v_{n}^{T}
[/eqn]

nice now I can shitpost irrelevant equations on sci with the rest of u faggots

[eqn]
\partial_\alpha(\sqrt{-\gamma}\gamma^{\alpha \beta}\partial_\beta X_\mu ) = 0
[/eqn]
ok

There is a neat way to prove this is true by counting.

nCn is 1 (0 when n=0), (n+1)Cn is (n+1).
(n+2)C(n+1) is (n+2).

There is no neat counting argument here, it is basic, especially if you exploit nCk = nC(n-k).

ah, my mistake 0C0 is also 1

...

nCn = n!/(n!0!) = n!/n!=1
(n+1)Cn = (n+1)!/(n!1!) = (n+1)!/n! = (n+1)n!/n! = n+1
nCn + (n+1)Cn = 1 + n+1 = n+2

(n+2)C(n+1) = (n+2)!/((n+1)!1!) = (n+2)!/(n+1)! = (n+2)(n+1)!/(n+1)! = n+2

=> nCn + (n+1)Cn = (n+2)C(n+1)

What do you think Pascal's triangle is

or C(n,n)=1 C(n+1,n)=n+1 and C(n+2,n+1)=n+2
dumbass