Question

Tính :

a ) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right)...\left(\frac{1}{99}+1\right)\)

b )\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{100}-1\right)\)

c ) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}\)

Answer 1

a)

\(=\frac{3}{2}.\frac{4}{3}......\frac{100}{99}=\frac{100}{2}=50\)

Answer 2

b)

\(=\frac{\left(-1\right)}{2}.\frac{\left(-2\right)}{3}.....\frac{\left(-99\right)}{100}=\frac{-1}{100}\)

Answer 3

c)

\(=\frac{\left(2^2-1\right)\left(3^2-1\right)......\left(30^3-1\right)}{2^2.3^2.....30^2}\)

\(=\frac{\left(2-1\right)\left(2+1\right)\left(3-1\right)\left(3+1\right)......\left(30-1\right)\left(30+1\right)}{2^2.3^2.....30^2}\)

\(=\frac{1.3.2.4.....29.31}{2^2.3^2.....30^2}=\frac{2.3^2.4^2.....29^2.30.31}{2^2.3^2.....30^2}=\frac{31}{2.30}=\frac{31}{60}\)