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Let $a_{k}=\binom{2n}{k}$,find the sum $$\dfrac{1}{a_{1}}-\dfrac{1}{a_{2}}+\dfrac{1}{a_{3}}-\dfrac{1}{a_{4}}+\cdots+\dfrac{1}{a_{2n-1}}-\dfrac{1}{a_{2n}}$$

$A:\dfrac{1}{n+1}$ $~~$ B:$-\dfrac{1}{n+1}~~~~~~~$C:$\dfrac{n}{n+1}~~~~$.D$-\dfrac{n}{n+1}$

I have use when $n=1,2$ get the answer $D$,But How to find this sum? Thanks

I consider $$\dfrac{(-1)^{k-1}}{a_{k}}=\dfrac{1}{f(k)}-\dfrac{1}{f(k+1)}$$,then we find the $f(k)?$

math110
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1 Answers1

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$$\sum_{k=1}^{n}\frac{(-1)^{k-1}k!(2n-k)!}{(2n)!}=\sum_{k=1}^{n}\frac{(-1)^{k-1}k\Gamma(k)\Gamma(2n-k+1)}{\Gamma(2n+1)}=\sum_{k=1}^{n}k(-1)^{k-1}B(k,2n-k+1)$$ and the RHS equals $$ \sum_{k=1}^{n}k(-1)^{k-1}\int_{0}^{1}(1-u)^{k-1}u^{2n-k}\,du=\int_{0}^{1}u^n\left(u^{n+1}-(-1)^n(1-u)^n(n+u)\right)\,du $$ or $$ \frac{1}{2n+2}-\frac{(-1)^n}{2\binom{2n}{n}}.$$

Jack D'Aurizio
  • 353,855