How to prove that $\sum_{k=0}^{\infty} \frac{1}{1+k} {9/2 \choose k}=\frac{2}{11}~(32\sqrt{2}-1)$

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How to prove that $$\sum_{k=0}^{\infty} \frac{1}{1+k} {9/2 \choose k}=\frac{2}{11}~(32\sqrt{2}-1).$$ I have no idea of handling binomial coefficient when the upper index is non-integer. Please help.

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$$(1+t)^x=\sum_{k=0}^{\infty}{x \choose k} t^k, 0<t \le 1, x>0$$ Integrating.w.r.t.$t$ from $t=0$ to $t=1$ both sides, we get $$\sum_{k=0}^{\infty} \frac{{x\choose k}}{k+1}=\frac{2^{x+1}-1}{x+1}$$ Put $x=9/2$, to get the required result.