Question

Tính A = \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{99.101}\right)\)

Answer 1

\(A=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{99.101}\right)\)

\(A=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}....\dfrac{10000}{99.101}\)

\(A=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}....\dfrac{100^2}{99.101}\)

\(A=\dfrac{2.3.4...100}{1.2.3....99}.\dfrac{2.3.4....100}{3.4.5....101}\)

\(A=100.\dfrac{2}{101}=\dfrac{200}{101}\)

Vậy A = \(\dfrac{200}{101}\)

Chúc học tốt!!