Câu hỏi của Trúc Giang

Question

$P=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{50^2}\right)$ So sánh P với $\frac{1}{2}$

Trần Minh Hoàng · Accepted Answer

Ta có:

$P=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{2500}\right)$

$P=\frac{3}{4}.\frac{8}{9}...\frac{2499}{2500}$

$P=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{49.51}{50.50}$

$P=\left(\frac{1}{2}.\frac{2}{3}...\frac{49}{50}\right).\left(\frac{3}{2}.\frac{4}{3}...\frac{51}{50}\right)$

$P=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}$Câu trả lời: Ta có:

$P=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{2500}\right)$

$P=\frac{3}{4}.\frac{8}{9}...\frac{2499}{2500}$

$P=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{49.51}{50.50}$

$P=\left(\frac{1}{2}.\frac{2}{3}...\frac{49}{50}\right).\left(\frac{3}{2}.\frac{4}{3}...\frac{51}{50}\right)$

$P=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}$

Bảo Hà · Answer

Ta có:

P=(1−14)(1−19)...(1−12500)P=(1−14)(1−19)...(1−12500)

P=34.89...24992500P=34.89...24992500

P=1.32.2.2.43.3...49.5150.50P=1.32.2.2.43.3...49.5150.50

P=(12.23...4950).(32.43...5150)P=(12.23...4950).(32.43...5150)

P=150.512=51100Câu trả lời: Ta có:

P=(1−14)(1−19)...(1−12500)P=(1−14)(1−19)...(1−12500)

P=34.89...24992500P=34.89...24992500

P=1.32.2.2.43.3...49.5150.50P=1.32.2.2.43.3...49.5150.50

P=(12.23...4950).(32.43...5150)P=(12.23...4950).(32.43...5150)

P=150.512=51100

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