Question

[Nguyen moc Huong]

Tinh bang cach hop li (1+1/3)× (1+1/8)× (1+1/15)×....× (1+1/9999) 

Answer 1

[Nguyễn Lê Phước Thịnh]

Ta có: \(\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{8}\right)\times...\times\left(1+\dfrac{1}{9999}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\times\left(1+\dfrac{1}{3^2-1}\right)\times....\times\left(1+\dfrac{1}{100^2-1}\right)\)

\(=\dfrac{2^2}{2^2-1}\times\dfrac{3^2}{3^2-1}\times...\times\dfrac{100^2}{100^2-1}\)

\(=\dfrac{2\times3\times...\times100}{1\times2\times...\times99}\times\dfrac{2\times3\times...\times100}{3\times4\times...\times101}=\dfrac{100}{1}\times\dfrac{2}{101}=\dfrac{200}{101}\)