Prove that ${2n \choose n} = {2n \choose n+1} + {2n \choose n-1}$

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?

• Sorry, I typed my equation wrong, going to fix it! – op_finales Apr 6 '16 at 23:22
• Try $n=2$ to try to show it's not true. – user84413 Apr 6 '16 at 23:28
• 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*} – Brian M. Scott Apr 6 '16 at 23:39
• What happened to your plan to fix the question, @op_finales ? – Thomas Andrews Apr 7 '16 at 0:02

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 ${4\choose 1}+{4\choose 3}=8$.
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 $n=1$.