use mathematical induction to prove the following formula for positive integer n.
@waterineyes
idk
its true for n=1 because \[\left(\begin{matrix}3 \\ 3\end{matrix}\right)=\left(\begin{matrix}4 \\ 4\end{matrix}\right)\]
now suppose that statement holds for \(n\) then we most prove its true for \(n+1\)
so we have \[\left(\begin{matrix}3 \\3\end{matrix}\right)+\left(\begin{matrix}4 \\3\end{matrix}\right)+...+\left(\begin{matrix}n+2 \\3\end{matrix}\right)=\left(\begin{matrix}n+3 \\4\end{matrix}\right)\] and we want to prove \[\left(\begin{matrix}3 \\3\end{matrix}\right)+\left(\begin{matrix}4 \\3\end{matrix}\right)+...+\left(\begin{matrix}n+3 \\3\end{matrix}\right)=\left(\begin{matrix}n+4 \\4\end{matrix}\right)\]
\[\left(\begin{matrix}3 \\3\end{matrix}\right)+\left(\begin{matrix}4 \\3\end{matrix}\right)+...+\left(\begin{matrix}n+3 \\3\end{matrix}\right)=\left(\begin{matrix}3 \\3\end{matrix}\right)+\left(\begin{matrix}4 \\3\end{matrix}\right)+...+\left(\begin{matrix}n+2 \\3\end{matrix}\right)+\left(\begin{matrix}n+3 \\3\end{matrix}\right)\\=\left(\begin{matrix}n+3 \\4\end{matrix}\right)+\left(\begin{matrix}n+3 \\3\end{matrix}\right)\]
so we just need to prove \[\left(\begin{matrix}n+3 \\4\end{matrix}\right)+\left(\begin{matrix}n+2 \\3\end{matrix}\right)=\left(\begin{matrix}n+4 \\4\end{matrix}\right)\]
am i clear?
yes what next ?
sorry my last reply is \[\left(\begin{matrix}n+3 \\4\end{matrix}\right)+\left(\begin{matrix}n+3 \\3\end{matrix}\right)=\left(\begin{matrix}n+4 \\4\end{matrix}\right) \]
now just note that... \[\left(\begin{matrix}n+3 \\4\end{matrix}\right)+\left(\begin{matrix}n+3 \\3\end{matrix}\right)=\frac{(n+3)!}{(n-1)! \ 4!}+\frac{(n+3)!}{n! \ 3!}=\frac{n(n+3)!+4(n+3)!}{n! \ 4!}=\frac{(n+4)!}{n! \ 4!}\\=\left(\begin{matrix}n+4 \\4\end{matrix}\right) \]
so the proof is complete and that statement holds for all \( n \in N\)
thanks for your help:)
yw........:)
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