Question

Let n and r be positive integers. Prove that.... ( n 0) + (n+1 1) + ... + (n + r r) = (n+r +1 r)

Answer 1

\[\left(\begin{matrix}n \\ 0\end{matrix}\right)+\left(\begin{matrix}n+1 \\ 1\end{matrix}\right)+....\left(\begin{matrix}n+r \\ r\end{matrix}\right)=\left(\begin{matrix}n+r+1 \\ r\end{matrix}\right)\]

Answer 2

thqat's what I'm trying to prove any ideas would be great it looks like pascal's formula

Answer 3

put \[\left(\begin{matrix}N+1 \\ n+1\end{matrix}\right)=\left(\begin{matrix}N \\ n\end{matrix}\right)+\left(\begin{matrix}N \\ n+1\end{matrix}\right)\]
so add and subtract \[\left(\begin{matrix}n+1 \\ 0\end{matrix}\right)\] and use first formula

Answer 4

sorry slightly confused where did the r go?

Answer 5

\[\left(\begin{matrix}n \\ 0\end{matrix}\right)+\left(\begin{matrix}n+1 \\ 1\end{matrix}\right)+....\left(\begin{matrix}n+r \\ r\end{matrix}\right)=\left(\begin{matrix}n+r+1 \\ r\end{matrix}\right)\]

\[{n+r+1\choose r}={n+r\choose r}+{n+r\choose r-1}\]
\[={n+r\choose r}+{n+r-1\choose r-1}+{n+r-1\choose r-2}\]
\[={n+r\choose r}+{n+r-1\choose r-1}+{n+r-2\choose r-2}+{n+r-2\choose r-3}=\cdots\]

Answer 6

in your form n(yours) =N and r =n (of mine)

Answer 7

Hey Zarkon I don't what you did in the first line how can \[\left(\begin{matrix}n+r+1 \\ r\end{matrix}\right)=\left(\begin{matrix}n+r \\ r\end{matrix}\right)+\left(\begin{matrix}n+r\\ r-1\end{matrix}\right)\] do you just add the top like normal and the bottom stuff like normal?

Answer 8

\[\left(\begin{matrix}N+1 \\ n+1\end{matrix}\right)=\left(\begin{matrix}N \\ n+1\end{matrix}\right)+\left(\begin{matrix}N \\ n\end{matrix}\right)\]

\(N=n+r\) and \(n+1=r\)

gives

\[\left(\begin{matrix}n+r+1 \\ r\end{matrix}\right)=\left(\begin{matrix}n+r \\ r\end{matrix}\right)+\left(\begin{matrix}n+r\\ r-1\end{matrix}\right)\]

Answer 9

oh I get it now thanks for your help!