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Question Number 120920 by Khalmohmmad last updated on 04/Nov/20

Answered by MJS_new last updated on 04/Nov/20

$${20}^{111}$$$$$$$$0\mathrm{because}k=0\Rightarrow k^2+k=0\mathrm{and}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{multiplying}$$$$$$$$2^{49}{d}^{49}=562949953421312d^{49}$$$$$$$$\underset{r=a}{\prod ^b}(2r+2k)=2^{b-a+1}\times \frac{(k+a)(k+b)!}{(k+a)!}$$$$\mathrm{for}a<0\mathrm{we}\mathrm{get}\mathrm{some}x!\mathrm{with}x\in {\mathbb{Z}}^{-}\mathrm{and}\mathrm{both}$$$$\mathrm{faculty}\mathrm{and}\mathrm{\Gamma }\left(x\right)\mathrm{are}\mathrm{not}\mathrm{defined}\mathrm{for}x\in {\mathbb{Z}}^{-}$$

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