Question

Tính A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{98^2}-1\right)\left(\frac{1}{99^2}-1\right)\)

Hi vọng các CTV và giáo viên olm giúp

Answer 1

A=\(\left(\frac{1}{2^2}-1\right)\)\(\left(\frac{1}{3^2}-1\right)\)\(\left(\frac{1}{4^2}-1\right)\)...\(\left(\frac{1}{98^2}-1\right)\)\(\left(\frac{1}{99^2}-1\right)\)

Do tích A có(99-2)+1=98 thừa số nguyên âm nên tích A dương

A=\(\frac{3}{4}\).\(\frac{8}{9}\).\(\frac{15}{16}\)...\(\frac{97.99}{98^2}\).\(\frac{98.100}{99^2}\)=\(\frac{1.2.3.4.5...97.98.99.100}{2^2.3^3.4^2...98^2.99^2}\)

=\(\frac{1.2.3.4...98}{2.3.4...98.99}.\frac{3.4.5...99.100}{2.3.4...98.99}=\frac{1}{99}.\frac{100}{2}=\frac{50}{99}\)