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Exercitiu gimnaziu: (1-1 supra 2)×(1-1 supra 3)×(1-1 supra 5)×...×(1-1 supra 50)

Răspuns :

Rayzen
[tex]P = \Big(1-\dfrac{1}{2}\Big)\cdot \Big(1-\dfrac{1}{3}\Big)\cdot\Big(1-\dfrac{1}{4}\Big)\cdot ...\cdot \Big(1-\dfrac{1}{50}\Big) \\ \\ P = \prod\limits_{k=2}^{50}\Big(1-\dfrac{1}{k}\Big) \\ P = \prod\limits_{k=2}^{50}\dfrac{k-1}{k} \\ P = \dfrac{\prod\limits_{k=2}^{50}(k-1)}{\prod\limits_{k=2}^{50}k} \\ P = \dfrac{\prod\limits_{k=2}^{50}(k-1)}{\prod\limits_{k=3}^{50}(k-1)\cdot 50} \\ \\ P = \dfrac{(2-1)\cdot \prod\limits_{k=3}^{50}(k-1)}{\prod\limits_{k=3}^{50}(k-1)\cdot 50}[/tex]
[tex] \\ \\ P = \dfrac{(2-1)}{50} \Rightarrow \boxed{\boxed{P = \dfrac{1}{50}}}[/tex]
[tex](1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})...(1-\frac{1}{50})=\\
\\ =(\frac{2}{2}-\frac{1}{2})(\frac{3}{3}-\frac{1}{3})(\frac{4}{4}-\frac{1}{4})...(\frac{50}{50}-\frac{1}{50})=\\
\\ =\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*...*\frac{49}{50}=\\
\\ =\frac{1}{50}[/tex] (restul se reduc)