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Mathematics 22 Online
OpenStudy (anonymous):

use mathematical induction to prove the following formula for positive integer n.

OpenStudy (anonymous):

OpenStudy (anonymous):

@waterineyes

OpenStudy (anonymous):

idk

OpenStudy (anonymous):

its true for n=1 because \[\left(\begin{matrix}3 \\ 3\end{matrix}\right)=\left(\begin{matrix}4 \\ 4\end{matrix}\right)\]

OpenStudy (anonymous):

now suppose that statement holds for \(n\) then we most prove its true for \(n+1\)

OpenStudy (anonymous):

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 \[\left(\begin{matrix}3 \\3\end{matrix}\right)+\left(\begin{matrix}4 \\3\end{matrix}\right)+...+\left(\begin{matrix}n+3 \\3\end{matrix}\right)=\left(\begin{matrix}n+4 \\4\end{matrix}\right)\]

OpenStudy (anonymous):

\[\left(\begin{matrix}3 \\3\end{matrix}\right)+\left(\begin{matrix}4 \\3\end{matrix}\right)+...+\left(\begin{matrix}n+3 \\3\end{matrix}\right)=\left(\begin{matrix}3 \\3\end{matrix}\right)+\left(\begin{matrix}4 \\3\end{matrix}\right)+...+\left(\begin{matrix}n+2 \\3\end{matrix}\right)+\left(\begin{matrix}n+3 \\3\end{matrix}\right)\\=\left(\begin{matrix}n+3 \\4\end{matrix}\right)+\left(\begin{matrix}n+3 \\3\end{matrix}\right)\]

OpenStudy (anonymous):

so we just need to prove \[\left(\begin{matrix}n+3 \\4\end{matrix}\right)+\left(\begin{matrix}n+2 \\3\end{matrix}\right)=\left(\begin{matrix}n+4 \\4\end{matrix}\right)\]

OpenStudy (anonymous):

am i clear?

OpenStudy (anonymous):

yes what next ?

OpenStudy (anonymous):

sorry my last reply is \[\left(\begin{matrix}n+3 \\4\end{matrix}\right)+\left(\begin{matrix}n+3 \\3\end{matrix}\right)=\left(\begin{matrix}n+4 \\4\end{matrix}\right) \]

OpenStudy (anonymous):

now just note that... \[\left(\begin{matrix}n+3 \\4\end{matrix}\right)+\left(\begin{matrix}n+3 \\3\end{matrix}\right)=\frac{(n+3)!}{(n-1)! \ 4!}+\frac{(n+3)!}{n! \ 3!}=\frac{n(n+3)!+4(n+3)!}{n! \ 4!}=\frac{(n+4)!}{n! \ 4!}\\=\left(\begin{matrix}n+4 \\4\end{matrix}\right) \]

OpenStudy (anonymous):

so the proof is complete and that statement holds for all \( n \in N\)

OpenStudy (anonymous):

thanks for your help:)

OpenStudy (anonymous):

yw........:)

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