Question

\(A=\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{9999}\right)\)

So sánh A với 2

Answer 1

\(A=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot\cdot\cdot\dfrac{10000}{9999}\)

\(=\dfrac{2.2}{3}\cdot\dfrac{3.3}{2.4}\cdot\dfrac{4.4}{3.5}\cdot\cdot\cdot\dfrac{100.100}{99.101}\)

\(=\dfrac{2.100}{101}=\dfrac{200}{101}=1,9801...< 2\)

Answer 2

\(A=\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right).....\left(1+\dfrac{1}{9999}\right)\)

\(A=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}....\dfrac{10000}{9999}\)

\(A=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}......\dfrac{100.100}{99.101}\)

\(A=\dfrac{2.3.4.5.....100}{1.2.3.4......99}.\dfrac{2.3.4.5.....100}{3.4.5.....101}\)

\(A=\dfrac{2.100}{101}=\dfrac{200}{101}=1,9801.....\)

Ta thấy: \(1.9801....< 2\)

Vậy A < 2