Question

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

Answer 1

\(\frac{1}{2^2}-1=\frac{1-2^2}{2^2}=\frac{\left(1-2\right)\left(1+2\right)}{2^2}=-1.\frac{3}{2^2}\)

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

Đặt nguyên biểu thức là B , ta có :

\(B=\left[-1.\left(-2\right).\left(-3\right)...\left(-99\right)\right].\frac{3.4.5...101}{\left(2.3.4.5...100\right)^2}\)

\(B=-\left(1.2.3...99\right).\frac{3.4.5...101}{\left(2.3.4.5...100\right)^2}\)

B=\(\frac{-2.\left(3.4.5...99\right)^2.100.101}{2^2\left(3.4.5...99\right)^2.100^2}=\frac{-101}{200}\)