Question

[Nguyễn Hải]

(1/2^2-1)(1/3^2-1)(1/4^2-1).....(1/100^2-1)

Answer 1

[Akai Haruma]

Lời giải:

\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1\right)\\ =\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}....\frac{1-100^2}{100^2}\\ =\frac{(2^2-1)(3^2-1)....(100^2-1)}{2^2.3^2.4^2...100^2}\\ =\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)....(100-1)(100+1)}{(2.3.4...100)(2.3.4...100)}\)

\(=\frac{(2-1)(3-1)(4-1)...(100-1)}{2.3.4...100}.\frac{(2+1)(3+1)(4+1)...(100+1)}{2.3.4...100}\\ =\frac{1.2.3...99}{2.3.4..100}.\frac{3.4.5..101}{2.3.4...100}\\ =\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)