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Question Number 115135 by Algoritm last updated on 23/Sep/20

Commented by Dwaipayan Shikari last updated on 23/Sep/20

$$\frac{1}{51.52...100}=\frac{1.2.3..}{100!}=\frac{50!}{100!}$$$$1.3.5..9...=\frac{1.2.3.4...}{2^{100}.50!}$$$$\frac{100!}{2^{50}.50!}.\frac{50!}{100!}=2^{\mathrm{x}}$$$$\mathrm{x}=-50$$

Answered by Olaf last updated on 23/Sep/20

$$\mathrm{B})$$$$\mathrm{P}=\frac{\underset{k=0}{\prod ^{50}}(2k+1)}{\underset{k=51}{\prod ^{100}}k}$$$$\mathrm{P}=\frac{\underset{k=1}{\prod ^{k=50}}2k\times \underset{k=0}{\prod ^{49}}(2k+1)}{\underset{k=1}{\prod ^{50}}2k\times \underset{k=51}{\prod ^{100}}k}$$$$\mathrm{P}=\frac{100!}{2^{50}\underset{k=1}{\prod ^{50}}k\times \underset{k=51}{\prod ^{100}}k}$$$$\mathrm{P}=\frac{100!}{2^{50}100!}=2^{-50}\Rightarrow x=-50$$$$$$$$$$

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