Question

[Lê Ngọc Linh]

\(\left(\dfrac{1}{2^2}-1\right)\).\(\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right)\).....\(\left(\dfrac{1}{100^2}-1\right)\)

giúp mình nhanh với. tối mình phải nộp rùi huhuhu

Answer 1

[Tăng Ngọc Đạt]

\(\left(\dfrac{1}{2^2}-1\right)\times\left(\dfrac{1}{3^2-1}\right)\times\left(\dfrac{1}{4^2}-1\right)\times...\times\left(\dfrac{1}{100^2}-1\right)\)

\(=\dfrac{3}{2^2}\times\dfrac{8}{3^2}\times\dfrac{15}{4^2}\times...\times\dfrac{100^2-1}{100^2}\)

\(=\dfrac{1\times3}{2\times2}\times\dfrac{2\times4}{3\times3}\times\dfrac{3\times5}{4\times4}\times...\times\dfrac{99\times101}{100\times100}\)

\(=\dfrac{1\times2\times3\times...\times99}{2\times3\times4\times...\times100}\times\dfrac{3\times4\times5\times...\times101}{2\times3\times4\times...\times100}\)

\(=\dfrac{1}{100}\times\dfrac{101}{2}\)

\(=\dfrac{101}{200}\)

Answer 2

[Nguyễn Đăng Nhân]

\(\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

\(=\dfrac{-3}{4}\cdot\dfrac{-8}{3}\cdot...\cdot\dfrac{-9999}{10000}\)

\(=\dfrac{1\cdot\left(-3\right)}{2\cdot2}\cdot\dfrac{2\cdot\left(-4\right)}{3\cdot3}\cdot...\cdot\dfrac{99\cdot\left(-101\right)}{100\cdot100}\)

\(=\dfrac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\dfrac{\left(-3\right)\cdot\left(-4\right)\cdot...\cdot\left(-101\right)}{2\cdot3\cdot4\cdot...\cdot100}\)

Ở tử số phân số bên phải có số thừa số là: \(101-3+1=99\)

99 là số lẻ nên tử số vế phải sẽ cho ra số âm.

\(=\dfrac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot\left(-101\right)}{2\cdot3\cdot4\cdot...\cdot100}\)

\(=\dfrac{1\cdot\left(-101\right)}{100\cdot2}\)

\(=\dfrac{-101}{200}\)