Question

[Nhật Mai]

Giúp bài 2 từ câu C (bài 3 nếu bn thích) nhé

Answer 1

[Nguyễn Minh Anh]

\(C=\left(1-\frac{1}{5}\right).\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{100}\right)=\frac{4}{5}.\frac{5}{6}...\frac{99}{100}=\frac{4}{100}=\frac{1}{25}\)

\(D=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)...\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...0...\left(1-\frac{10}{7}\right)=0\)

\(E=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2021}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2020}{2021}=\frac{1}{2021}\)

\(G=\left(\frac{1-2^2}{2^2}\right).\left(\frac{1-3^2}{3^2}\right).\left(\frac{1-4^2}{4^2}\right)...\left(\frac{1-2021^2}{2021^2}\right)=\frac{\left(1-2\right)\left(1+2\right)}{2^2}.\frac{\left(1-3\right)\left(1+3\right)}{3^2}.\frac{\left(1-4\right)\left(1+4\right)}{4^2}...\frac{\left(1-2021\right)\left(1+2021\right)}{2021^2}=\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}...\frac{\left(-2020\right).2022}{2021^2}=\frac{3.4...2021.2022}{3.4...2021}.\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-2020\right)}{2^2.3.4...2021}=2022.\frac{1}{2.2021}=\frac{1011}{2021}\)