0
$\begingroup$

How can I prove this? Assume $n \ge 0$ integer.

$$ \frac {\binom{2(n+1)}{n+1}}{n+2} - \sum_{k=1}^n \left( \frac{4\binom{2(n-k)+1}{n-k-1}}{n-k+3} + \frac{\binom{2k}{k}}{k+1} \right) = 1.$$

  • 0
    It looks like there's a lot of basic simplifications that you can do... – 2017-02-27
  • 0
    Are you sure? When $k=n$, the term $\binom{2(n-k)+1}{n-k-1}$ has a $-1$ in the "denominator". How is $\binom{1}{-1}$ defined? – 2017-02-27
  • 0
    @JackD'Aurizio If k<0 or k>n then binomial(n,k) = 0. I think this is standard convention. $\binom{1}{-1} = 0$. – 2017-02-27

1 Answers 1

1

Assuming $\binom{1}{-1}=0$, the LHS equals $$ \frac{1}{n+2}\binom{2n+2}{n+1}-\sum_{k=1}^{n}\frac{1}{k+1}\binom{2k}{k}-\sum_{k=1}^{n-1}\frac{4}{k+3}\binom{2k+1}{k-1} $$ hence it is enough to check that the claim holds for $n\in\{0,1,2\}$, then check that $$ \frac{1}{n+2}\binom{2n+2}{n+1}-\frac{1}{n+1}\binom{2n}{n} = \frac{1}{n+1}\binom{2n}{n}+\frac{4}{n+2}\binom{2n-1}{n-2} $$ holds for every $n\geq 2$. The claim follows by induction.

  • 0
    Thanks Jack. Induction is the last rescue for formulas like these. I wait to see if Michael Burr shows the "lot of basic simplifications that you can do" which he sees, but I do not, which of course would be more interesting. – 2017-02-27