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I need to prove that $${1\over{n+1}} {2n \choose n} = {2n \choose n} - {2n \choose n-1}$$

I started by writing out all the terms using the formula ${n!\over{k!(n-k)!}}$ but I can't make the two sides equal.

Thanks for any help.

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    Dec 3, 2013 at 15:07

4 Answers 4

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$$ {2n \choose n}-{1\over{n+1}} {2n \choose n}$$

$$=\binom{2n}n\left(1-\frac1{n+1}\right)$$ $$=\frac{(2n)!}{n! n!}\cdot\frac n{n+1}$$ $$=\frac{2n!}{\{(n+1)\cdot n!\}\{n\cdot(n-1)!\}}\cdot n$$

$$=\frac{2n!}{(n+1)!\cdot (n-1)!}$$

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

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Because $(n+1)! = (n+1)n!$ and $n! = n(n-1)!$, we have $$ \binom{2n}{n-1} = \frac{(2n)!}{(n-1)!(n+1)!} = \frac{n}{n+1}\binom{2n}{n} $$ and the rest is easy.

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Multiply both sides by $n+1$. On RHS you get $\binom{2n}{n}(n-n+1)$ which will be equal to LHS. This is because $$ \binom{2n}{n-1}\cdot (n+1)=\binom{2n}{n-1}\cdot (n+1)\cdot \frac{n}{n}=n \cdot \binom{2n}{n} $$

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That should work. Note that $$ \begin {align*} \dbinom {2n}{n-1} &= \dfrac {(2n)!}{(n-1)! (n+1)!} \\&= \dfrac {n}{n+2} \cdot \dfrac {(2n!)}{n!n!} \\&= \dfrac {n}{n+1} \cdot \dbinom {2n}{n}, \end {align*} $$ so we have: $ \dbinom {2n}{n-1} = \left( 1 - \dfrac {1}{n+1} \right) \cdot \dbinom {2n}{n}, $ which is equivalent to what we wanted to show.

It seems I've been beaten by 2 answers . . .

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