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I am trying to prove that.

$\displaystyle {2n \choose n} = \displaystyle {2n \choose n+1} + {2n \choose n-1}$

Is this necessarily true / how should I try to tackle it if it is?

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  • $\begingroup$ Sorry, I typed my equation wrong, going to fix it! $\endgroup$ – op_finales Apr 6 '16 at 23:22
  • $\begingroup$ Try $n=2$ to try to show it's not true. $\endgroup$ – user84413 Apr 6 '16 at 23:28
  • $\begingroup$ As others have noted, what you have is not generally true. In fact, $$\begin{align*}\color{red}{\binom{2n+2}{n+1}}-2\binom{2n}n &=\color{red}{\binom{2n+1}{n+1}+\binom{2n+1}n}- 2\binom{2n}n\\ &=\color{blue}{\binom{2n+1}{n+1}}-\binom{2n}n+\color{green}{\binom{2n-1}n}- \binom{2n}n\\ &=\color{blue}{\binom{2n}{n+1}+\binom{2n}n}-\binom{2n}n+\color{green}{\binom{2n}n+\binom{2n}{n-1}}-\binom{2n}n\\ &=\binom{2n}{n+1}+\binom{2n}{n-1}\;. \end{align*}$$ $\endgroup$ – Brian M. Scott Apr 6 '16 at 23:39
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    $\begingroup$ What happened to your plan to fix the question, @op_finales ? $\endgroup$ – Thomas Andrews Apr 7 '16 at 0:02
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It's not true in general that ${2n\choose n}={2n\choose n-1}+{2n\choose n+1}$. For instance, if $n=2$ then ${4\choose 2}=6$, while ${4\choose 1}+{4\choose 3}=8$.

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In general, $$ \binom{2n}{n-1}=\binom{2n}{n+1}=\frac{n}{n+1}\binom{2n}{n} $$ Therefore, $$ \frac{\binom{2n}{n-1}+\binom{2n}{n+1}}{\binom{2n}{n}}=\frac{2n}{n+1} $$ which is $1$ if and only if $n=1$.

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